confidence limits for the range of a search radar

Confidence Limits for the Range of a Search RadarAuthor(s): Colin StevensSource: Journal of the Royal Statistical Society. Series C (Applied Statistics), Vol. 6, No. 3(Nov., 1957), pp. 214-222Published by: Wiley for the Royal Statistical SocietyStable URL: http://www.jstor.org/stable/2985607 .

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CONFIDENCE LIMITS FOR THE RANGE OF A SEARCH RADAR

COLIN STEVENS

Communication from the Staff of the Applied Electronics Laboratories of the General Electric Company Ltd, Stanmore, England

The determination of confidence limits for the percentage points of a normal distribution is a problem with an exact, but somewhat troublesome, solution. In this article Mr Stevens describes a fairly simple approximate solution, which was found to be sufficiently accurate.

THE function of a search radar is the detection of distant targets, and in general the greater the range of detection the better the radar may be said to be. Detection range is therefore an important aspect of search radar performance, and this paper is concerned with its determination from data obtained from trials.

There are, however, certain difficulties involved in the definition of radar range. In the first place it varies with the nature and aspect of the target, since a large target reflects more energy than a small one, and the reflecting area of a given target also varies considerably with its aspect. It is therefore customary to specify the nature and aspect of the target in some such phrase as 'Mosquito target, head-on'. Secondly, in the case of rapidly approaching targets, it has long been recognised that there is no such thing as 'the range' of a search radar, but rather a distribution of detection ranges. A number of causes contribute to this effect, including random variations in the strength of the received echo, the presence of unwanted and spurious random echoes on the display, variability of operators, and day-to-day variations in the setting up of the radar. It is essential to take this variability into account when defining detection range, and the definition in current use is framed in terms of the 'ac%0 detection range' (ac being some number between 0 and 100) defined as that range at or by which a% of incoming targets are detected. This definition is illustrated in Fig. 1, in which the upper figure shows the definition in terms of the probability density function (or frequency distribution) of detection ranges, and the lower in terms of the cumulative probability function. A numerical value of 75 is used for oc in some radar specifications. Using x to denote the detection range (a random variable) and X<x to denote the a% detection range, the definition of X< can be concisely expressed as

Pr[x > X zc 0~c~).... (1) x (O < a < 1)

21I4



RADAR CONFIDENCE LIMITS 215

The problem with which this paper is concerned is the determination of the xc% detection range from experimental data. Specifically, given a random sample of N observed detection ranges, xl, x2, ..., 2XN, what can be said about X,, ? The natural approach is to plot the cumulative distribution of the data and thus make a direct estimate of X<Z. This

LZ

Qz

00

RANGE (x)

F-

0

U Q\

00 LU.

Oca

-0

0 a-

0

xx RANGE (x)

FIG. 1. Probability distribution of detection ranges; the upper diagram shows the probability density function and the lower diagram shows the cumulative probability function.

process is illustrated in Fig. 2, where a straight line has been fitted by eye to some hypothetical data plotted on normal probability paper; the data serve as the basis of a worked example discussed later in this paper. This graphical method has the advantage of dispensing with distributional assumptions. (A straight line has been fitted on normal probability paper in Fig. 2 because the data are known in this case to be normally distributed. The method may in principle be applied equally well to a smooth curve drawn through the same points plotted on ordinary squared paper.)

However, it is not of much use merely to estimate X< without giving some indication of the degree of sampling fluctuation to which the estimate is liable. This fluctuation arises from the fact that X, is



2I6 APPLIED STATISTICS

estimated from the data of the sample, and if the sampling process were repeated a number of times one would expect the samples to yield differing estimates of X,:.

12-

11- - - - - -- - -

LU~ ~ ~~~~~~~~e z

z__ o 10 9 9 - - 40 - -

U FI Estimated 75r/o Detection be asserted tlbteXa Xwta dsep aiRange=9y4

99 98 95 90 80 60 40 20 10 5 2 1

CUMULATIVE PROBABILITY (%)

FiG. 2. Cumulative distribution of sample data on detection ranges plotted on normal probability graph paper.

A convenient way of expressing the precision of an estimate is by the statement of a confidence interval, defined by a lower limit X1 aild an upper limit X2, in such a manner that the unknown quantity X,, can be asserted to lie between Xi and X2 with any desired probability y that the assertion is correct. It should be remembered that X"' is not a random variable but an unknown constant of the distribution; the limits X1 and X2 are random variables. The statement really means that in long series of experiments a proportion y of intervals (X1, X2) computed from each experiment would contain X,, and a proportion 1 - y would not.

To determine X1 and X2 we need the sample mean x and standard deviation s defined by

XR =(Xl + X2 + * X XN +..)/. (2)

(N - l)s2 = (X1 - X)2 + (X2 - X)2 + . . . + (XN - . . . . (3)

We also require K6, defined as the standardised normal deviate ex- ceeded with probability E. If t is normal with zero mean and unit variance, K, is defined by

Pr[t K> K]j (O < < 1) * (4)



RADAR CONFIDENCE LIMITS 2 I 7

Values of IC may be obtained from standard tables of the normal integral: as examples, K0.025 71960 0, K0.25 0674 5, K.o975

0-025 -1X960 0. The problem may be solved exactly by means of the non-central

t distribution, and the method for doing this, with the necessary tables, is given by Johnson and Welch' (see Example 3 on page 374). The exact method, however, requires a troublesome iterative calculation, and interpolation in the tables. The present method, which is approximate, requires only the solution of a quadratic equation, and, for samples of size ten or more, yields solutions accurate enough for most practical purposes. A comparison of results obtained by the two methods is given in the Appendix for two examples, in each of which the sample size is ten. In neither case does the width of the confidence interval obtained by the approximate method differ b-y more than 2% from the exact value.

The basis of the approximation, that the variate x + ks is normally distributed with mean ft + ka and variance g2[/IN + k2/2(N - 1)], is discussed in some detail by Johnson and Welch, and is shown to be a very good approximation for the present purpose.

The method now to be described is based on the assumption that the observations are independent random samples from a normal population of unknown mean and standard deviation. The limited amount of data available to me suggests that this is a reasonable assumption to make, and, from a theoretical point of view, the number of independent random processes affecting detection range (external and internal 'noise', operators, and variations in the tune of the set) could also be expected to lead to an approximately normal distribution of results.

The method is given in step-by-step form.

STEP 1. Compute x and s from the sample data.

STEP 2. Compute _ (1 - y)/2 .... (5) where y is the probability to be attached to the confidence interval.

STEP 3. Obtain K,, and K, from tables of the normal integral.

STEP 4. Solve the quadratic equation in k:

(i1- 2(ji1l)) k2 k 21Mk + {K2- 0 ...(6)

Let the roots of this equation be k1 and k2, k, being the smaller.

STEP 5. The confidence limits Xl, X2 are then given by

Xi, + kis'17 X2X + k2S

Numerical Example The methods described will be illustrated by application to a hypo-

thetical sample of detection ranges, obtained from a table of random



2I8 APPLIED STATISTICS

normal deviates with mean 10 and standard deviation 1. The sample data are as follows:

95 9-8 94 8-4 94 10-2 11-4 9-2 11-7 10.9 10-2 8-7 10*5 9-2 10*1 9-9 9-3 10-5 10*9 10-8

Suppose we are required to estimate the 75 % detection range (cc 0.75). For the purpose of making a preliminary investigation the data have been plotted on normal probability paper in Fig. 2, by ordering the data in the form xl > x2 > .. . > xi > ... > XN, and plotting the ith ordered observation xi at the position [ 100 (i - 1) /N]00. The use of these plotting positions is discussed by Chernhoff and Liebermann,2 who show that they yield a quite efficient estimate of the population standard deviation, and hence presumably an efficient estimate of a percentage point of the distribution. By means of a line fitted by eye to the data, the 7500 detection range can be estimated as 9-4. This happens to be in good agreement with the true 7500 detection range of the population from which the sample was drawn, i.e. 10-00 - 0-67 9.33.

Suppose we are now required to determine the 9500 confidence interval for the 750% detection range, i.e. y 0 95.

STEP 1 X 1000 s 0-880 8

(That x is identical with the population mean is pure coincidence.) STEP 2 - (1 - 0 95)/2 0-025 STEP 3 KW K0.75 =-K0.25 -0X674 5

K=a K0.025 1960 0 STEP 4. Since in this case N= 20, the quadratic equation in k

becomes (0.898 91)k2 + 2(0-674 5)k + 0-262 87 = 0

yielding the roots kl 1-271 k2 =-0-230

STEP 5. The required confidence limits are therefore X1 = 10*00 - (1-271)(0-880 8) = 8-88 X2= 1000 - (0 230)(0 880 8) 9-80

We therefore make the assertion, with a 9500 chance of being right, that the 750% detection range of the radar lies between 8-88 and 9-80.

Number of Observations for an Interval of Given Width The result of (6) may be used to determine the width W of the

confidence interval, defined by W = IXI X21 .... (8)

From (7): W =k -- k2is .. . .(9)



RADAR CONFIDENCE LIMITS 2I9

By solving (6), or otherwise, we can obtain an expression for Ik- k2- , and (9) may then be written

(N?2 Koc2 - 'Ii = 2 N 2(N-1) 2Nf(N-1) ....(10)

2(N - 1) This expression is valuable as a guide to the number of observations required to establish a confidence interval of predetermined width (e.g. 10 miles) with any given degree of confidence y. This can only be done, however, if a reliable estimate of is available beforehand which can be used for s in (1 0). Another feature of this result is that, for large N, the width W varies inversely as the square root of N, so that, for example, four times the number of observations are required to reduce the width of the interval to a half.

The ratio W/s has been plotted in Fig. 3 for 95 0 confidence intervals for three different values of a, 0-75, 0 95, 0.99.

2

0 _____0_99

2 0 \d=~~~~~0 952

10 1 0 20 40 60 100

NUMBER OF OBSERVATIONS (N)

FIG. 3. Variation of width of 95% confidence limits for X, with number of observations. W = width of confidence interval; s = sample standard deviation; y 095.

Example Suppose we have to estimate beforehand the number of observations

required to determine the 75%0 detection range to an accuracy of 1 unit with 95%0 confidence. Previous experience suggests a value for a of 1-7 units. Then W ~ 1, and we take W/s 1/1.7 0 59. Reading across the figure on to the c = 0 75 curve, we obtain N - 61. This then is our estimate of the number of observations required.

In estimating W47 by these means it should be remembered that its accuracy depends on the accuracy of the original estimate of a, and also that u has been used for s in the calculation, thereby making no allow- ance for fluctuations in the sample standard deviation s.



220 APPLIED STATISTICS

APPENDIX

Let x denote detection range, and let Xa denote the range such that

Pr[x>Xj = ....(A1)

The problem is to determine limits X1, X2 (X1 < X2), computed from a sample of N observations in such a manner that the interval (X1, X2) can be asserted to contain the point Xx with probability y that the assertion is correct. Since there is an infinite number of such intervals, we shall add the further restriction that we seek a central confidence interval, i.e. one satisfying the following equations:

Pr[(- oo, X1) contains X (1 - y)/2 -q .... (A 2)

Pr[(X2, oo) contains Xj (1 - y)/2 -q .... (A 3)

in which (- co, X1) denotes the interval with - co and X1 as its end points, etc.

It is assumed that the variate x is normally distributed with unknown mean Mt and unknown standard deviation a. In consequence of this we have

X. =M~u + K,,,,g ....(A4)

This leads us to investigate functions of the type

Xi x + kis (i= 1, 2) ....(A 5)

for the limits X1, X2. The Xi are statistical variates, being functions of the sample values, and it is therefore legitimate to write (A 2), (A 3) as

Pr[Xj > X=] =J .... (A 6)

Pr[X2<XJ = .... (A7)

The first of these can be written in the form

Pr[x + k1s > M + K,] 7g

which can be rearranged as follows:

Pr (N) x /o - K.\IN > -klN aX . * (A 8) S//

This establishes the connection with non-central t, since this is defined as

t ~ (z + 5) /V s/e.... (A 9)

in which z is normal with zero mean and unit variance, and co is independently distributed as x2/f, f being the degrees of freedom of the x2. The point t(f, 5, E) is defined as the number such that

Pr[t > t(f, 5, E)|f, 3] = E

and it follows therefore that

-k1l'N= t( - 1, -I VN, 7\ ) ... . .(A 10)



RADAR CONFIDENCE LIMITS 22I

The tables of non-central t permit this equation to be solved exactly for kl, and k2 can be found in a similar manner. In practice, however, this can be a troublesome process, and we shall therefore seek an approximate solution for k1 and k2 which can be found in a simple manner.

Approximate Solution for Con?fidence Limits Consider X1 = x + kis as a linear combination of two variates x

and s. Under the normality assumed for the observations the following remarks may be made:

(i) x and s are independent. (ii) x is normally distributed with mean M and variance u2/N. (iii) The distribution of s tends to normality with increasing N, with

mean and variance given by

E(s) au ....(A 11)

Var(s) b2U2/2(N - 1) .... (A 12)

The constants a and b depend on N, and approach unity as N increases. For example, when N = 10, a = 0973, b 0-985. We shall hereafter assume that a = b = 1, an assumption discussed by Johnson and Welch' and shown to lead to a more accurate solution of the present problem than that obtained by giving them their correct values.

(iv) Since x is normal and s nearly normal, the linear combination x + ks is also nearly normal, and more so than s.

We shall therefore assume Xi to be normal, with mean and variance given by

E (Xi) -M + kia .... (A 13)

Var(Xi) I2/N + k 22/2(N - 1) .... (A 14)

Now consider (A 6):

Pr[Xl > Xj = Pr (X - E (X1) > XO -E (Xj) ) uX YX,

and therefore X- E(X1) K-ax, .... (A 15)

From (A 4), (A 13), (A 14) we have

K.- k [1/N + kl12/2(N 1)] ... (A 16)

A similar argument applied to (A 7) yields, for k2:



222 APPLIED STATISTICS

By squaring both sides of (A 16), (A 17) we see that k, and k2 are roots of a quadratic equation in k:

[1 - K2/2(N - 1)]k2 - 2KXk + [I2 - 2/N] - 0 .... (A 18) which is the required result.

A Note on the Accuracy of the Approximation In order to evaluate the accuracy of the approximation, exact and

approximate solutions of two examples have been computed for comparison.

CASE I NJ 10,x 075, y 095

CASE II NJ 10, Xa 010, Y 090 (This is Example 3 in Johnson and Welch's paper.')

The exact and approximate solutions are as follows:

CASE I Exact Approximate

k, -1*673 ki -1661

k2 -0*056 k2 -0054

|k- k21 1*617 Ik1 -k2 1*607

CASE II Exact Approximate

ki= 0*712 k1 0*696 k2= 2*355 k2 2*321

I - k2 1 1*643 I - k2 1 1*625

REFERENCES 1 JOHNSON, N. L., and WELCH, B. L. (1940). 'Applications of the non-central t

distribution', Biomet7ika, 31, 362. 2 CHERNHOFF, H., and LIEBERMANN, G. J. (1954). 'Use of normal probability paper',

_7. Amer. Statist. Ass., 49, 778.



confidence limits for the range of a search radar

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