bfn/bfn - technical university of denmark · imm - dtu 02405 probability 2005-10-29 bfn/bfn the...
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![Page 1: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/1.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is
binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 2: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/2.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) =
312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 3: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/3.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 4: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/4.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) =
31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 5: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/5.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212=
34. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 6: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/6.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4.
The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 7: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/7.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =
3∑i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 8: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/8.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8=
3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 9: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/9.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 10: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/10.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the
computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 11: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/11.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 12: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/12.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) +
(E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 13: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/13.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =
3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 14: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/14.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 15: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/15.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4=
3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 16: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/16.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 17: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/17.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 18: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/18.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 19: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/19.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) =
3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 20: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/20.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) =
3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 21: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/21.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 22: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/22.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 23: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/23.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) =
E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 24: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/24.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)−
(E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 25: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/25.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 26: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/26.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) =
E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 27: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/27.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =
3 + 16 · 3 + 81 · 18
=132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 28: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/28.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 29: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/29.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 30: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/30.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =
33
2− 9 =
15
2
![Page 31: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/31.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2
![Page 32: BFN/bfn - Technical University of Denmark · IMM - DTU 02405 Probability 2005-10-29 BFN/bfn The random variable Y is binominally distributed. The mean E(Y) = 31 2 = 3 2, the variance](https://reader031.vdocument.in/reader031/viewer/2022020304/5b9d3a6209d3f29a298c21bb/html5/thumbnails/32.jpg)
IMM - DTU 02405 Probability
2005-10-29BFN/bfn
The random variable Y is binominally distributed. The mean E(Y ) = 312= 3
2, the variance is
V (Y ) = 31212= 3
4. The Mean of Y 2 is
E(Y 2) =3∑
i=0
i2(3
i
)(1
2
)i(1
2
)3−i
= 13
8+ 4
3
8+ 9
1
8= 3
Alternatively we could have used the computational formula for the variance
V (Y ) + (E(Y ))2 =3
4+
9
4= 3
P (Y = 0) =
(1
2
)3
P (Y = 1) = 3
(1
2
)3
P (Y = 2) = 3
(1
2
)3
P (Y = 3) =
(1
2
)3
The variance of Y 2 can be found by
V (Y 2) = E((Y 2)2)− (E(Y 2))2
We need to calculate E((Y 2)2) = E(Y 4)
E(Y 4) =3 + 16 · 3 + 81 · 1
8=
132
8=
33
2
Finally
V (Y 2) =33
2− 9 =
15
2