Variance of max of $m$ i.i.d. random variables

extreme-value-analysisextreme-value-theoremprobability theoryvariance

I'm trying to verify if my analysis is correct or not.

Suppose we have $m$ random variables $x_i$ , $i \in m$. Each $x_i \sim \mathcal{N}(0,\sigma^2)$.

From extreme value theorem one can state $Y= \max\limits_{i \in m} [\mathcal{P}(x_i \leq \epsilon)] = [G(\epsilon)]$ as $m\to\infty$, if $x_i$ are i.i.d and $G(\epsilon)$ is a standard Gumbel distribution.

My first question is can we state that: $$\text{Var}[Y]= \text{Var}\left[\max_{i \in m} [\mathcal{P}(x_i \leq \epsilon)] \right]= \text{Var}[ [G(\epsilon)]] = \frac{\pi^2}{6}$$

My second question is, if we have $n$ of such $Y$ but all of them are independent with zero mean, can we state:
$$\text{Var}\left[\prod_{i}^n Y_i\right] = \left(\frac{\pi^2}{6}\right)^n$$

Thanks.

Update:
There's final result for the second point at Distribution of the maximum of a large number of normally distributed random variables but no complete step by step derivation.

Best Answer

$\def\dto{\stackrel{\mathrm{d}}{→}}\def\peq{\mathrel{\phantom{=}}{}}$The answers to question 1 and 2 are both negative.

For question 1, since $Y_m = \max\limits_{i \in m} P(x_i \leqslant ε) \dto G(ε)$, then $\color{blue}{\lim\limits_{m → ∞}} D(Y_m) = D(G(ε))$, i.e. the equality is in the sense of a limiting process.

For question 2, it's geneally not true that$$ D\left( \prod_{m = 1}^n Y_m \right) = \prod_{m = 1}^n D(Y_m) $$ for i.i.d. $Y_1, \cdots, Y_n$, especially when $E(Y_1) ≠ 0$ and $D(Y_1) > 0$ by the following proposition:

Proposition: If $X$ and $Y$ are independent random variables on the same probability space, then$$ D(XY) - D(X) D(Y) = D(X) (E(Y))^2 + D(Y) (E(X))^2. $$

Proof: Since $X$ and $Y$ are independent, then\begin{gather*} D(XY) = E(X^2 Y^2) - (E(XY))^2 = E(X^2) E(Y^2) - (E(X) E(Y))^2,\\ D(X) D(Y) = \left( E(X^2) - (E(X))^2 \right) \left( E(Y^2) - (E(Y))^2 \right)\\ = E(X^2) E(Y^2) - E(X^2) (E(Y))^2 - E(Y^2) (E(X))^2 + (E(X))^2 (E(Y))^2, \end{gather*} and\begin{align*} &\peq D(XY) - D(X) D(Y) = E(X^2) (E(Y))^2 + E(Y^2) (E(X))^2 - 2 (E(X))^2 (E(Y))^2\\ &= \left( E(X^2) - (E(X))^2 \right) (E(Y))^2 + \left( E(Y^2) - (E(Y))^2 \right) (E(X))^2\\ &= D(X) (E(Y))^2 + D(Y) (E(X))^2. \tag*{$\square$} \end{align*}

Now it can be proved by induction on $n$ with the above proposition that$$ D\left( \prod_{m = 1}^n Y_m \right) > \prod_{m = 1}^n D(Y_m) > 0. $$

Related Solutions

[Math] Distribution of sum of multiplication of i.i.d. exponential random variables.

In answer to your first question ...

Given $X \sim Exponential(\lambda_1)$ with $E[X] =\lambda_1 $, and $Y \sim Exponential(\lambda_2)$ with $E[Y] =\lambda_2 $, where $X$ and $Y$ are independent. Let:

$$W_i =c (X_i-a) (Y_i-b) \quad \text{and} \quad Z_n = \sum_{i=1}^n W_i$$

Then, by the Lindeberg-Levy version of the Central Limit Theorem:

$$Z_n\overset{a} {\sim }N\big( n E[W], n Var(W)\big)$$

We immediately have: $$E[W] = c \left(\lambda _1-a\right) \left(\lambda _2-b\right)$$

Variance of $W$

The OP attempts to approximate the variance - this is not necessary and causes errors.

By independence, the joint pdf of $(X,Y)$ is $f(x,y)$:

Then, $Var[W]$ is:

where I am using the Var function from the mathStatica package for Mathematica to do the nitty-gritties. All done.

Central Limit Theorem approximation

Here are $100000$ pseudo-random drawings of $Z$ generated in Mathematica, given $n = 200, \lambda_1= 3, \lambda_2 =2,a=2.2,b=4$ ...

zdata = Table[
     xdata = RandomVariate[ExponentialDistribution[1/3], {200}];
     ydata = RandomVariate[ExponentialDistribution[1/2], {200}];
     Total @@ {(xdata - 2.2) (ydata - 4)}, {i, 1, 100000}];

The CLT Normal approximation $N\big(\mu, \sigma^2\big)$ has parameters $\mu = n E[W]$ and $\sigma = \sqrt{n Var(W)}$:

Here, the squiggly BLUE curve is the empirical pdf (from the Monte Carlo data), and the dashed red curve is the Central Limit Theorem Normal approximation. It works very nicely WHEN THE CORRECT variance derivation is used, even with a sample of size $n = 200$.

Central Limit Theorem fit using OP's approximated variance

By contrast, if we use the OP's approximation of Var(Z) to calculate $\sigma$, then the CLT 'fit' is not good at all:

Notes

As disclosure, I should add that I am one of the authors of the software used above.

Maximal type inequality for sum (or average) of i.i.d. random variables

For people who are interested, this question has been answered here on mathoverflow by fedja.

Best Answer

Related Solutions

[Math] Distribution of sum of multiplication of i.i.d. exponential random variables.

Maximal type inequality for sum (or average) of i.i.d. random variables

Related Question