1
$\begingroup$

Hi I am still having some trouble with the following question: I have mostly figured out the first part but after that is where I get confused

Say we have random variables $X $~$ Poisson(\lambda)$ and $\lambda_{1} \gt \lambda_{0} \gt 0$ set values and we want to test $H_{o}: \lambda=\lambda_{o}$ and $H_{1}: \lambda=\lambda_{1}$.

The question asks to show that the optimal test at a level $\alpha$ rejects the null hypothesis when $\bar X_n \gt c$ and find $c$, where $\bar X_n=\frac{1}{n}(x_{1}+...+x_{n})$ and furthermore show that the test that minimizes the sum of type one and type two errors rejects the null hypothesis when $\bar X_n \gt c*$ and find $c*$

What I have tried:

I simply use that the optimal test will reject the null hypothesis when $$f(x|\lambda_{o}) \lt kf(x|\lambda_{1})$$

ie when $$\frac{f(x|\lambda_{1})}{f(x|\lambda_{o})} \gt \frac{1}{k}$$

for which I solve that $$\bar X_{n} \gt \frac{-lnk+n(\lambda_{1}-\lambda_{o})}{nln(\lambda_{1}/\lambda_{o})}$$

So if I call the RHS my c then our optimal test rejects the null when $\bar X_{n} \gt c$

Now I know I also want to have that $Pr(\bar X_{n} \gt c : \lambda=\lambda_{o})=\alpha$ that is I want the c such that $Pr(\bar X_{n} \gt c) $given that $\lambda=\lambda_{o}$ is $\alpha$.

Ps: Is it correct to say that the optimal test at level alpha will be that test such that the type 2 error is smallest for alpha type one error? and I am confused on how to do that,

would it be like $$\sum_{x=0}^{c'}\frac{e^{{-\lambda_{o}} \lambda_{o}^{x}}}{x!}=\alpha$$ I mean how can I find such c?

But now I am confused on the second part. I want to show that the test that minimizes the sum of the type 1 and type 2 errors rejects the null when $\bar X_{n} \gt c*$ and find that $c*$.

I know that type 1 error occurs when we reject the null even though it is true and type 2 occurs when we accept the null even though it is false. So I am confused on this part. I know our optimal test rejected in the conditions above at level $\alpha$ so is it the same test

type 1:

$P(Xn \gt c : H_{o})=\alpha$

Type 2

$P(Xn \le c : H_{1})=\beta$

I have been trying for very long and will greatly appreciate any help one can offer, I am just confused on putting it all together. Thanks

$\endgroup$
0
$\begingroup$

I'll give my opinion on this matter. The aim is indeed to compute $c$ such that $\mathbb{P}(\overline{X}_n > c \ | \ \lambda = \lambda_0) = \alpha$. To that end you'll first need to verify the distribution of $\overline{X}_n$. It is well known (and easy to prove) that the sum of independent Poisson variables is again Poisson distributed. More precisely, \begin{align} &\mathbb{P}(\overline{X}_n > c \ | \ \lambda = \lambda_0) \\ & = \mathbb{P}(X_1+\ldots+X_n > nc \ |\ \lambda = \lambda_0) \\ & = \sum_{k = nc+1}^{\infty} \frac{(n\lambda_0)^k e^{-n \lambda_0}}{k!} \\ & = e^{-n \lambda_0} \left(\sum_{k=0}^{\infty} \frac{(n \lambda_0)^k}{k!} -\sum_{k=0}^{nc} \frac{(n \lambda_0)^k}{k!} \right) \\ & = e^{-n \lambda_0} \left(e^{n \lambda_0} - \underbrace{\sum_{k=0}^{nc} \frac{(n \lambda_0)^k}{k!}}_{(*)} \right) = \alpha \end{align} There is a closed form formula for the partial sum $(*)$ of the exponential series that might be useful, however it can turn out in an implicit expression for $c$.

$\endgroup$
  • $\begingroup$ Thanks yes so I know that they can be added , but then what is the final answer , what is that you have equal to alpha ? $\endgroup$ – Quality Apr 2 '16 at 23:44
  • $\begingroup$ Perhaps you could use an approximated value, see for instance: math.stackexchange.com/questions/136996/…. What also might work is to approximate the Poisson distribution by a normal distribution (assuming $n$ is large enough). In that case (under the null hypothesis) $\overline{X}_n \sim N(\lambda_0, \lambda_0/n)$. This wil allow you an easier calculation of $\mathbb{P}(\overline{X}_n > c \ | \ \lambda = \lambda_0)$. $\endgroup$ – Cedric Cavents Apr 3 '16 at 6:46
  • $\begingroup$ "There is a closed form formula..." > There is NO closed form formula... $\endgroup$ – Did Apr 4 '16 at 5:35
  • $\begingroup$ Apparently, I used the wrong terminology. I meant that it is not possible to find an explicit expression for $c$ (as far as I know). $\endgroup$ – Cedric Cavents Apr 4 '16 at 10:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.