The i.i.d. random variables X_1 and X_2 are Poisson(theta): that is, they have p

Question

The i.i.d. random variables X_1 and X_2 are Poisson(theta): that is, they have probability distribution P(X_i = k)= e^-theta (theta^k/k!), k = 0, 1, 2, ... We consider the two (simple) hypotheses H_0: theta = 1 and H_1: theta = 2. Define: f(k_1, k_2|H_i) = P(X_1 = k_1, X_2 = k_2|H_i), with i = 0, 1. We consider the test IF f(X_1, X_2 |H_0)/f(X_1, X_2|H_1) > 2 THEN accept H_0, ELSE reject H_0. (a) Show that the ratio only f(X_1, X_2 |H_0)/f(X_1, X_2|H_1) depends on the statistic S = X_1 + X_2. What is the acceptance region R_0 of the above test? That is, for which values of S do we accept H_0? (b) Find the significance level alpha of this test. (c) Find the power 1 - alpha of this test.

Explanation / Answer

Solution

Back-up Theory

If X and Y are iid, then P(X = x, Y = y) = P(X = x) x P(Y = y) …………………. (1)

Let 1 and 2 be respectively the value of under H0 and H1.

Then, the ratio, R say, = {f(x1, x2/H0)}/{f(x1, x2/H1)}

The numerator of R = e- 1(1)x1 /(x1)! x e- 1(1)x2 /(x2)! [Theory (1) above]

= e- 21(1)x1 + x2/{(x1)! x (x2)!}

Similarly, denominator of R = e- 22(2)x1 + x2/{(x1)! x (x2)!}

Thus, R = e- 2(1 - 2) (1/2)x1 + x2 ……………………………..……………….(2)

Part (a)

Given 1 = 1 and 2 = 2, [vide (2) above], R = e2 (½)x1 + x2 ………………….(3)

e2 being a constant, (3) => R depends only on the statistic S = x1 + x2 First part DONE

The given stipulation of accepting H0 for R > 2 => e2 (½)x1 + x2 > 2 [vide (3) above]

=> [by taking log with base e], 2 + (x1 + x2)ln(½) > ln2 or (x1 + x2) > (ln2 – 2)/ln(½)

or (x1 + x2) > (0.6931 – 2)/(- 0.6931) = 1.6656.

So, we accept H0, if S > 1.6656 ANSWER

Part (b)

Level of significance = P(H0 is rejected when H0 is true)

= P{(x1 + x2) < 1.6656/ = 1}

= [P(X1 = 0, X2 = 0) + P(X1 = 0, X2 = 1) + P(X1 = 1, X2 = 0) with = 1]

= (e– 1 x e– 1) + (e– 1 x e– 1) + (e– 1 x e– 1) = 3e– 2 = 3 x 0.1353 = 0.4059 ANSWER

Part (c)

Power of the test = 1 - P(H0 is accepted when H1 is true)

= 1 – [1 - P(H0 is rejected when H1 is true)]

= P(H0 is rejected when H1 is true)

= P{(x1 + x2) < 1.6656/ = 2}

= [P(X1 = 0, X2 = 0) + P(X1 = 0, X2 = 1) + P(X1 = 1, X2 = 0) with = 1]

= (e– 2 x e– 2) + (e– 2 x 2e– 2) + (2e– 2 x e– 2) = 5e– 4 = 5 x 0.0183 = 0.0915 ANSWER

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