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Let $X_1,X_2,\ldots,X_N$ be i.i.d random variables having $\text{Exp}(1)$ distribution where $N$ is unknown. Suppose only $T=\max\{X_1,X_2,\ldots,X_N\}$ is observed.

I have to derive a most powerful test for testing $H_0:N=5$ versus $H_1:N=10$.

So $N$ is my parameter of interest. The joint distribution of $X_1,\ldots,X_N$ is of the form

$$f_N(x_1,\ldots,x_N)=\exp\left({-\sum_{i=1}^N x_i}\right)\mathbf1_{{x_1,\ldots,x_N>0}}$$

But this cannot be my likelihood function since $x_1,\ldots,x_N$ is not observed.

I am not sure how to express the above joint density as a function of $t$, the observed value of $T$. If I can write the joint density as some $f_N(t)$, then that would be my likelihood function given $t$.

I understand that the test is of the form $$\varphi(t)=\mathbf1_{\lambda(t)>k}$$

, where $\lambda$ is the likelihood ratio $$\lambda(t)=\frac{f_{H_1}(t)}{f_{H_0}(t)}$$

Any hint would be much appreciated.

As an aside to the actual question, I am curious if it is possible to find MLE of $N$.

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  • $\begingroup$ Your joint distribution does not depend on the individual $x_i:$ it depends only on their sum, which is observed. $\endgroup$ – whuber Dec 21 '18 at 16:09
  • $\begingroup$ @whuber Yes it depends only on the sum. So $0<x_1,\ldots,x_N\le t\implies 0<\sum_{i=1}^N x_i\le Nt$. Does this imply $\sum_{i=1}^N x_i$ is observed? It only bounds the sum. $\endgroup$ – StubbornAtom Dec 21 '18 at 16:25
  • $\begingroup$ There's a missing step here: under your model assumptions, you can easily derive the distribution of $T.$ $\endgroup$ – whuber Dec 21 '18 at 16:49
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    $\begingroup$ @whuber Somehow did not realize that I am supposed to use the distribution of $T$. That makes it a standard problem. $\endgroup$ – StubbornAtom Dec 21 '18 at 17:28
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Hint: the likelihood for $T$ is $$ f_{T}(t ; N) = N(1-e^{-t})^{N-1}e^{-t}. $$ To verify this, first find the cdf of $T$, and then differentiate. \begin{align*} F_T(t) &= P(T \le t)\\ &= [F_{X_i}(t)]^N. \end{align*}

Regarding your other question as to whether there is an MLE estimate for this: yes there is, but the likelihood is only defined on $\mathbb{N}^{+}$. This means that you cannot take a derivative and set it equal to zero.

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As an aside to the actual question, I am curious if it is possible to find MLE of $N$.

This is quite straightforward, and simply requires you to derive the log-likelihood function and then use standard (discrete or continuous) calculus techniques to maximise. Using rules for the distribution of order statistics, the sampling distribution for $T$ is:

$$f_T(t) = N e^{-t} (1-e^{-t})^{N-1}.$$

The log-likelihood is:

$$\ell_t(N) = \ln N - t + (N-1)\ln(1-e^{-t}).$$

Now, maximising this function can be done either by using discrete calculus (i.e., using difference operators), or it can be done by treating $N$ as continuous and maximising using continuous calculus, and then discretising the result. For simplicity, we will do the latter. Taking $N$ as a real variable, the log-likelihood has corresponding score function and Fisher information (essentially the first and second derivatives, but with the sign reversed on the second) given by:

$$\begin{equation} \begin{aligned} s_t(N) &\equiv \frac{d \ell_t}{dN}(N) = \frac{1}{N} + \ln(1-e^{-t}), \\[12pt] I_t(N) &\equiv - \frac{d^2 \ell_t}{dN^2}(N) = \frac{1}{N^2} >0. \\[12pt] \end{aligned} \end{equation}$$

Since $I_t(N) > 0$ the likelihood function is a strictly concave function, so this means it has a unique MLE at its sole critical point $s_t(\hat{N}(t)) = 0$. This gives you the continuous MLE:

$$\hat{N}(t) = \frac{1}{\ln(1-e^{-t})}.$$

This will generally not be an integer value, so you obtain the corresponding discrete MLE by looking at the two integers either side of this value; the bigger one is the discrete MLE (which is almost surely unique).

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If $N=5$ then $\Pr(X_1 \le x\ \&\ \cdots \ \&\ X_N\le x) = \left( 1-e^{-x} \right)^5,$ so you have a density function $$ \frac d {dx} \left( \left( 1-e^{-x} \right)^5 \right) = 5\left( 1 - e^{-x} \right)^4 \cdot e^{-x}. $$ And similarly if $N=10.$ So the likelihood function is $$ \begin{cases} L(5) = 5(1-e^{-x})^4 \cdot e^{-x} \\[8pt] L(10) = 10(1-e^{-x})^9 \cdot e^{-x} \end{cases} $$ where $x$ is the observed maximum value, and the ratio is $$ \frac{L(5)}{L(10)} = \frac 1 {2(1-e^{-x})^5}. $$ A small value of this ratio favors the alternative hypothesis $N=10.$ Equivalently, a large value of the observed maximum favors the alternative. Given the probability distribution of the maximum assuming the null hypothesis $N=5,$ you can find the critical value as a function of the level of the test.

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