Question

Consider the Poisson probability distribution, $p(x) = frac{e^{-lambda}lambda^x}{x!}$ for $x = 0, 1, 2, ...$ a) The score function, $U(lambda)$, is defined as the partial derivative of the logarithm of $p(x)$ with respect to $lambda$. Find $U(lambda)$. b) Find the expected value of $U(lambda)$ and the variance of $U(lambda)$. c) Given a random sample of size $n$ from $p(x)$, the sum is a Poisson random variable with parameter $nlambda$. Suppose $n = 8$. Suppose the null hypothesis that $lambda = 0.5$ is rejected in favor of the alternative that $lambda > 0.5$ if the sum of the random sample values is greater than or equal to 8. Find the power of the test if $lambda = 1$.

          Consider the Poisson probability distribution, $p(x) = frac{e^{-lambda}lambda^x}{x!}$ for $x = 0, 1, 2, ...$
a) The score function, $U(lambda)$, is defined as the partial derivative of the logarithm of $p(x)$ with respect to $lambda$. Find $U(lambda)$. 
b) Find the expected value of $U(lambda)$ and the variance of $U(lambda)$. 
c) Given a random sample of size $n$ from $p(x)$, the sum is a Poisson random variable with parameter $nlambda$. Suppose $n = 8$. Suppose the null hypothesis that $lambda = 0.5$ is rejected in favor of the alternative that $lambda > 0.5$ if the sum of the random sample values is greater than or equal to 8. Find the power of the test if $lambda = 1$.

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$Consider the Poisson probability distribution, p(x) = frace^-lambdalambda^xx! for x = 0, 1, 2, ... a) The score function, U(lambda), is defined as the partial derivative of the logarithm of p(x) with respect to lambda. Find U(lambda). b) Find the expected value of U(lambda) and the variance of U(lambda). c) Given a random sample of size n from p(x), the sum is a Poisson random variable with parameter nlambda. Suppose n = 8. Suppose the null hypothesis that lambda = 0.5 is rejected in favor of the alternative that lambda > 0.5 if the sum of the random sample values is greater than or equal to 8. Find the power of the test if lambda = 1.$

Added by Jessica S.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Madhur L

10/06/2023

Step 1

So, let's first find the logarithm of $p(x)$: $$ \ln p(x) = -\lambda + x\ln\lambda - \ln x! $$ Now, let's find the partial derivative with respect to $\lambda$: $$ U(\lambda) = \frac{\partial \ln p(x)}{\partial \lambda} = -1 + \frac{x}{\lambda} $$ Show more…

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Consider the Poisson probability distribution, p(x) = e^(-λ) * (λ^x) / x!, for x = 0, 1, 2, ... a) The score function, U(λ), is defined as the partial derivative of the logarithm of p(x) with respect to λ. Find U(λ). b) Find the expected value of U(λ) and the variance of U(λ). c) Given a random sample of size n from p(x), the sum is a Poisson random variable with parameter nλ. Suppose n = 8. Suppose the null hypothesis that λ = 0.5 is rejected in favor of the alternative that λ > 0.5 if the sum of the random sample values is greater than or equal to 8. Find the power of the test if λ = 1.

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Transcript

00:01 Hello students, here x follows poisson lambda, then pmf of x is p of x is equal to e raised to minus lambda lambda raised to x by x factorial where x is equal to 0, 1, 2, 3 etc.

00:14 In a part of the question, log p of x can be written as log e raised to minus lambda lambda raised to x by x factorial.

00:28 Therefore, log p of x becomes minus lambda plus x log lambda minus log x factorial.

00:41 Now, let us take the differentiation of log p of x with respect to lambda.

00:47 Dou by dou lambda log p of x can be written as dou by dou lambda of minus lambda plus x log lambda minus log x factorial is equal to minus 1 plus x by lambda.

01:15 Therefore, u of lambda is equal to x by lambda minus 1.

01:22 In b part of the question, we want to find expectation of u of lambda and variance of u of lambda.

01:29 Expectation of u of lambda can be written as expectation of x minus lambda minus 1.

01:38 This will become expectation of x by lambda minus 1 is equal to 1 by lambda into expectation of x minus 1.

01:56 X follows poisson lambda.

01:59 Therefore, expectation of x is equal to lambda and variance of x is equal to lambda.

02:09 Therefore, expectation of u of lambda becomes 1 by lambda into lambda minus 1 is equal to 1 minus 1.

02:21 Therefore, expectation of u of lambda is equal to 0.

02:29 Now, let us find variance of u of lambda.

02:33 Variance of u of lambda is variance of u of lambda can be written as variance of x by lambda minus 1 is equal to variance of x by lambda.

02:55 This can be written as while taking constant outside, it will become squared.

03:02 1 divided by lambda square into variance of x is equal to 1 by lambda square into variance of x is lambda which is 1 by lambda...

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