Mean and variance of the maximum of a random number of Uniform variables

Question

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 $M = \text{max}(U_i)$

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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.

Answer 1

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

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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.

Answer 2

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 $f(m|Y=y)$:

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

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

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

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

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