For the asymptotic distribution of the Kolmogorv-Smirnov statistic D, the latest NR edition gives two series. One, with terms ~exp(-n^2 D^2), is useful for large D. Another, with terms ~exp(-n^2 / D^2), is useful for small D.

For the asymptotic distribution of the Kuiper statistic V, NR only gives only one series, useful for large V. Can anyone point me to a corresponding formula useful for small V?

For the asymptotic distribution of the Kolmogorv-Smirnov statistic D, the latest NR edition gives two series. One, with terms ~exp(-n^2 D^2), is useful for large D. Another, with terms ~exp(-n^2 / D^2), is useful for small D.

For the asymptotic distribution of the Kuiper statistic V, NR only gives only one series, useful for large V. Can anyone point me to a corresponding formula useful for small V?


Hello NR Comrade ,


There is a simple definition Kuiper's statistc defined as
V= ( D+) + ( D-) = max[Sn(x)-P(x)] + max [P(x)-Sn(x)]
where limit of x tends from -infinity to +infinity
is the sum of th max distance of Sn(x) above and below P(x)
D refers integration.
The subject formula is to be tried under the circle .Sketch the indefinite integral of two probability distribution defined on the the circle as function of angle around the circle as, the angle goes through several times 360 degrees.

"Trying is equal to Winning"

Thanking You
As
Kutta(C.R.Muthukumar)
:)

That's a formula for computing V given an EDF. It's not a formula for the PDF of V under the null hypothesis.

That's a formula for computing V given an EDF. It's not a formula for the PDF of V under the null hypothesis.

Hello NR Comrade,
Please obtain a solution:->
when a die was thrown 9000 tiimes and of these,3220 yielded a 3 or 4. Is this consistent with hypothesis that the die was unbiasesd.
On the samelines why not for the above .

Bye for now-more later
Thanking You
As
Kutta(C.R.Muthukumar)

Hello NR Comrade,
Please obtain a solution:->
when a die was thrown 9000 tiimes and of these,3220 yielded a 3 or 4. Is this consistent with hypothesis that the die was unbiasesd.
On the samelines why not for the above .

Bye for now-more later
Thanking You
As
Kutta(C.R.Muthukumar)

Hello NR Comrade,
In coninuation of the above i have placed a solu for the same for ur needful try further
The Procedure is as follows:
Let getting a 3 or 4 be treated as a success;
Given that n =9000
P =observed portion of success = 3220/9000;
= 0.358;
NullHypothesis H(0): P != 1/3 ie) dice was biased;
Alternatve Hypothesis H(1): P != 1/3(two tailed);

Test Statistic::
Since n is large H(0) the test statistic is given by
Z= (p-P)/(SE of p) == p - P/sqrt.( PQ/n) == (0,1)
== 0.358 - 0.333/sqrt((1/3 *2/3)/9000) == 0.025/(sqrt.0.000025)=0.25/0.005
== 5;
SignificanceLevel ( from infin to ==) 0.05;
Critical Region:->
Since alternative hypothesis is two taled we shall apply twotailed
test for testing significance of Z
The critical value of Z at 5 % level of significance for two tailed test is 1.96
The calculated value of Z at 5 % level of significamce is 5
The calculted value of Z(5) is greater than the critial value (1.96) at 5% level of sinificance
Hence the null Hypothesis is rejected at 5% level of significance ie)
The dice was not biased for the dice was biased.

Hope this is for grasp .So Please let me know if y 'r able to get your required formula for smaller values of n;
Thanking You
As
Kutta(C.R.Muthukumar)

kutta, what you just posted has nothing whatever to do with the question being discussed in this thread. (And it contains mistakes; for instance "the die is biased" is not a sensible null hypothesis.)

I've been able to solve my own problem, and give the solution here for anyone who's interested:

P = \frac{\sqrt{2 \pi}}{x^3} \sum_{n=1}^{\infty} \pi^2 n^2 e^{-\pi^2 n^2 / 2 x^2}

Here's how to get it. Start with the formula in NR for the distribution of V.

Q = 2 \sum_{n=1}^{\infty} (4 n^2 x^2 - 1) e^{-2 n^2 x^2}

Express Q using the Jacobi theta function

g = \sum_{n=1}^{\infty} e^{-2 n^2 x^2}

and its derivative g'. Then use the identity relating the Jacobi theta function for one nome to the inverse nome. (This is the same identity used to derive the relationship between P and Q for the KS D statisitic.) This gives a formula for the same quantity as a series in e^{-n^2 / x^2} instead of e^{n^2 x^2}, which is just our formula for P above.

I've coded this formula and verified that they agree in the transition region where both series converge.