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Let $U \sim \text{Uniform}(0,1)$, and let $(U_1, U_2, \dots, U_Y)$ denote an iid sample of size $Y$, where the number of drawings $Y$ is itself a random variable with pmf:

$$P(Y=y)=\frac{1}{(e-1)y!} \quad \text{ for } y = 1, 2 \dots$$

Here, $Y$ is also independent of $U$.

My interest is to find the expected value and variance of $M$, where $M = \text{max}(U_i)$


Here is what I have so far:

I started with defining the cdf of M as follows.

$Pr(U1<=M)=\int_0^M{U1dU1}=\frac{M^2}{2}$

Since the cdf is the same for all iid $Ui's$,

$Pr(M<=m)=(\frac{m^2}{2})^y$

Is this correct? If so, how do I proceed from here? Can I treat y as a constant and find the pdf of M by integrating the cdf?

Also, I computed for $E[Y]=\sum_{y=1}^\infty {\frac{y}{(e-1)y!}}=\frac{1}{(e-1)}+1$ but I'm not yet sure how/if this can help.

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I have never met your exercice before, but this one is interesting and leads to funny calculation.


First, let me say that your computation are not totally correct. In fact you have computed the law of the random variable M | Y=y. Here are a complete answer.

Let one determine the pdf of the random variable M | Y=y. I emphasise that here, Y is fixed. It is easy to see that \begin{align} \mathrm{p}(M \leq x | Y=y) =& p(U_1, \ldots,U_Y \leq x | Y=y) \\ =& \prod_{y=1}^y p(U_1 \leq x) \\ =& \prod_{y=1}^y \int_0^x 1 \mathrm{d}t \\ =& x^y \\ =& \int_0^x yt^{y-1} \mathrm{d}t \end{align} So the pdf of $M|Y=y$ is the function $t\rightarrow yt^{y-1}$.

To find the pdf of $M$, a simple chain rule is enough. Hence \begin{align} \mathrm{p}(M \leq x) =& \sum_{y=1}^{+\infty} \mathrm{p}(M\leq x, Y=y) \\ =& \sum_{y=1}^{+\infty} \mathrm{p}(M\leq x | Y=y) \mathrm{p}(Y=y) \\ =& \int_0^x \sum_{y=1}^{+\infty} \frac{1}{(e-1)y!} yt^{y-1} \mathrm{d}t \\ =& \int_0^x \sum_{y=1}^{+\infty} \frac{1}{(e-1) (y-1)!} t^{y-1} \mathrm{d}t \\ =& \int_0^x \frac{e^t}{e-1} \mathrm{d}t. \end{align}

As a consequence, the pdf of $M$ is the function $t \rightarrow e^t / (e-1)$. From the pdf of M, it is easy to derive the two first moments of the distribution. Two IPP lead to \begin{align} \mathbb{E}[M] =& \frac{1}{e-1} \\ \mathrm{Var}(M) =& \frac{e^2-3e+1}{(e-1)^2} \end{align}

Here it is. I hope I have answered your question.

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  • $\begingroup$ We obtain the same pdf for $M$, but your variance calculation $\mathrm{Var}(M) =\frac{e-4}{(e-1)^2}$ appears to be incorrect: note that since $e<4$, your variance would be negative. $\endgroup$ – wolfies Jul 2 '17 at 15:09
  • $\begingroup$ Thank you @wolfies to point out this error. I have corrected my post accordingly. $\endgroup$ – TheCatInTheClock Jul 2 '17 at 15:20
  • $\begingroup$ Thank you. The insight that I indeed only had the cdf of M|Y was what I was missing. $\endgroup$ – user164144 Jul 2 '17 at 21:34
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Here is quick derivation of the steps using a computer algebra system (mathStatica with Mathematica).

By familiar arguments, the pdf of the maximum of $y$ standard Uniform random variables $M = \text{max}(U_1, \dots, U_y)$ is say $f(m|Y=y)$:

enter image description here

The sample size $Y$ is, instead of being fixed, itself a random variable with given pmf say $g(y)$:

enter image description here

The parameter mix distribution $E_g(f)$ is say $h(m)$:

enter image description here

with domain of support $M$ defined on $(0,1)$.

Finally, we seek $E[M]$ and $\text{Var}(M)$:

enter image description here

Note that the solution for $\text{Var}(M)$ is different to that posted above as $\mathrm{Var}(M) =\frac{e-4}{(e-1)^2}$.

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