Distribution of maximum of i.i.d. Chi-Square random variables with degree-of-freedom 2

Question

I wondering how the probability distribution of the maximum of i.i.d. Chi-square (two degrees-of-freedom) random variables $X_i \sim \chi^2(2)$ has the same distribution as $\sum_{i=1}^K \frac{X_i}{i}$, i.e., $$\max_{i=1,\ldots,K} X_i \ \stackrel{d}{=} \ \sum_{i=1}^K \frac{X_i}i .$$

Can anyone help me to prove this? Thank you

Answer 1

Let $X_{(1)},\ldots,X_{(K)}$ be the order statistics.

Let $Y_1 = X_{(1)}$ and for $i=2,\ldots,K$ let $Y_i= X_{(i)} - X_{(i-1)}.$

Then $\displaystyle \max \{ X_1,\ldots,X_n\} = \sum_{i=1}^K Y_i.$

If you can show that $Y_1,\ldots,Y_K$ are independent and $Y_i$ is exponentially distributed with expected value $2/i$ then you've got it.

For that you need the memorylessness of the exponential distribution.