# Sum of exponential random variables over their indices

Let $\ X_1,X_2,...\$ be i.i.d. exponential random variables. For $\ n=1,2,...\$consider the random variables $\ Y_n= \max\{X_1,...,X_n\} \$ and $\ U_n=\sum_{i=1}^{n}X_i/i \$. Show that for each n, the random variables $\ Y_n \$and $\ U_n \$ have the same distribution function.

I know that the sum of exponentials is a Gamma(n,$\lambda$) distribution, but the indices in the denominator here are making it difficult. In addition, I know the distribution function of $\ Y_n\$is $\ (1-\exp^{-\lambda x})^n\$. So I just need to show the distribution of $\ U_n$ is the same.

• – NCh Jan 2 '18 at 15:51

You are right that the distribution of $\ Y_n= \max\{X_1,...,X_n\} \$ goes something like this

$$P(Y_n \le x) = F_1(x).F_2(x).... F_n(x) = \left(F_i(x)\right)^n = (1-e^{-\lambda x})^n$$

$$f_{Y_n}(x) = n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}\tag 1$$

Now, let $\frac{X_1}{1} = Z_1$, $\frac{X_2}{2} = Z_2$, ... $\frac{X_n}{n} = Z_n$

$$P(X_1 \le x) = P(Z_1\le x) = (1-e^{-\lambda x})$$

$$f_{Z_1}(x) = \lambda.e^{-\lambda x}$$

Similarly

$$P(X_2 \le 2x) = P(Z_2\le x) = (1-e^{-2\lambda x})$$

$$f_{Z_2}(x) = 2\lambda.e^{-2\lambda x}$$

And thus

$$P(X_n \le nx) = P(Z_n\le x) = (1-e^{-n\lambda x})$$

$$f_{Z_n}(x) = n\lambda.e^{-n\lambda x}$$

So $Z_1$ ~ $Exp(\lambda)$, $Z_2$ ~ $Exp(2\lambda)$, .., $Z_n$ ~ $Exp(n\lambda)$

Refer to the article below to find the distribution of sum of exponentials with different parameters to get $U_n = \sum \frac{X_i}{i} = Z_1+Z_2+\cdots+Z_n$

https://people.maths.bris.ac.uk/~mb13434/sumexp.pdf

$$f_{Z_1+Z_2+\cdots+Z_n}(x) = \left(\prod_{i=1}^{n} \lambda_i\right)\sum_{j=1}^{n}\dfrac{e^{-\lambda_j x}}{\prod_{k\ne j k=1}^{n}(\lambda_k-\lambda_j)}\tag 3$$

$$=n\lambda\left( e^{-\lambda x} - (n-1).e^{-2\lambda x}+ \frac{(n-1)(n-2)}{2}e^{-3\lambda x} + \cdots + {(n-1)\choose(n-1)}e^{-n\lambda x}\right)$$

$$=n\lambda e^{-\lambda x}\left( 1- (n-1).e^{-\lambda x}+ \frac{(n-1)(n-2)}{2}e^{-2\lambda x} + \cdots + {(n-1)\choose(n-1)}e^{-(n-1)\lambda x}\right)$$

$$f_{U_n}(X) = f_{Z_1+Z_2+\cdots+Z_n}(x)= n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}\tag 2$$

Hence we have proved that $Y_n$ and $U_n$ have the same distribution. To illustrate the first few terms of $U_n$ in (3),

j = 1,

$$=(1.2.3...n)\lambda^n \dfrac{e^{-\lambda x}}{(2\lambda - 1\lambda) (3\lambda - 1\lambda)(4\lambda - 1\lambda)..(n\lambda - 1\lambda)}$$ $$=(1.2.3...n)\lambda^n \dfrac{e^{-\lambda x}}{1.2.3.4..(n-1)\lambda^{n-1}}$$

$$= n\lambda.e^{-\lambda x}$$

j = 2

$$=(1.2.3...n)\lambda^n \dfrac{e^{-2\lambda x}}{(1\lambda - 2\lambda) (3\lambda - 2\lambda)(4\lambda - 2\lambda)..(n\lambda - 2\lambda)}$$ $$=(1.2.3...n)\lambda^n \dfrac{e^{-2\lambda x}}{-1.1.2.3.4..(n-2)\lambda^{n-1}}$$

$$= -n(n-1)\lambda.e^{-2\lambda x}$$

j = 3

$$=(1.2.3...n)\lambda^n \dfrac{e^{-3\lambda x}}{(1\lambda - 3\lambda) (2\lambda - 3\lambda)(4\lambda - 3\lambda)..(n\lambda - 3\lambda)}$$ $$=(1.2.3...n)\lambda^n \dfrac{e^{-3\lambda x}}{-2.-1.1.2.3.4..(n-3)\lambda^{n-1}}$$

$$= \frac{n(n-1)(n-2)}{2}\lambda.e^{-3\lambda x}$$