Examining $\Pr(Y_u - X_u > z)$ where $X_u,Y_u$ are sample means of independent exponential variables

Question

I am trying to compare these two collections. Specifically, I am interested in examining $\Pr(Y_u - X_u > z)$ where $X_u$ and $Y_u$ denote the sample means of each. We can assume $X_i$ and $Y_i$ are all independent of each other

Here is what I was working on:

If we have both $X, Y \sim \operatorname{Exp}(\lambda)$ then we can make the joint distribution: $\frac{1}{\lambda^2} e^{-\frac{x}{\lambda}}e^{-\frac{y}{\lambda}}$. Let's say $Z = X - Y$.

From here we say

\begin{align} F_Z(z) &= 1 - P(Z > z) \\&= P(Y \geq z) \\& = 1 - \int_z^\infty \int_0^{x-z}\frac{1}{\lambda^2} e^{-\frac{x}{\lambda}}e^{-\frac{y}{\lambda}} \,dy\, dx \\& = 1 - \frac{1}{2}e^{-\frac{z}{\lambda}} \end{align}

If I plug in some values though this doesn't seem to represent what I had initially set out to find. Let me know if I need to add more details.

EDIT: If the sample size is large enough can this be approximated through some application of CLT?

Answer 1

The sample mean of $X_1, \ldots, X_n$, which you denote $X_u$, is gamma distributed, namely $$X_u \sim \operatorname{Gamma}(n, n\lambda), \\ f_{X_u}(x) = \frac{(n \lambda)^n x^{n-1} e^{-\lambda x/n}}{\Gamma(n)}, \quad x > 0.$$ Similarly, $$Y_u \sim \operatorname{Gamma}(n, n\lambda).$$ Their difference, $Z = Y_u - X_u$, is known as a variance-gamma distribution which does not have an elementary closed form for either the density or survival; the density, however, is expressible in terms of Bessel functions. Specifically, $$Z \sim \operatorname{VarianceGamma}(\mu = 0, \lambda^* = n, \alpha = n\lambda, \beta = 0),$$ where $\mu$ is the location parameter, $\lambda^*$ is a shape parameter (not to be confused with $\lambda$, but corresponds to $\lambda$ in the Wikipedia article I linked to), $\alpha$ is another shape parameter, and $\beta$ is a skewness parameter. For $n$ sufficiently large, the shape of this density begins to approximate a normal distribution with mean $0$ and variance $2/(n \lambda^2)$.

Answer 2

First of all we have

$$P(Y - X \leq z) = P(Y \leq z+X)$$

This is

$$ = \int_0^\infty P(Y \leq z+X) \mid X=x]f_X(x) dx $$ $$ = \int_0^\infty P(Y \leq z+x)\cdot f_X(x) dx = \int_0^\infty F_Y(z+x) f_X(x) dx , \, \forall \ z>0$$

$$ = \int_0^\infty (1- e^{-\lambda(z+x)}) \lambda e^{-\lambda x} dx = \lambda \int_0^\infty e^{-\lambda x} dx - \lambda e^{-\lambda z} \int_0^\infty e^{-(\lambda+\lambda)x} dx $$ $$ = \lambda \left[ \frac{e^{-\lambda x}}{-\lambda} \right]^\infty_0 - \lambda e^{-\lambda z} \left[ \frac{e^{-(2\lambda)x}}{-(2\lambda)} \right]^\infty_0 =1 - \frac{\lambda e^{-\lambda z}}{2\lambda} =1 - \frac{ e^{- \lambda z}}{2} $$

So $P(Y-X>z)=(Z>z)=...$

Answer 3

The pdf for a mean of $n$ independent exponential random variables all with parameter $\lambda$ is

$$\frac{n^n \lambda ^n y^{n-1} e^{-n\lambda y}}{(n-1)!}$$

for $0<y<\infty$ and $0$ elsewhere. To derive $\text{Pr}(Y-X>z)$ one integrates the product of the pdf's for $X$ and $Y$ over the appropriate values.

For $z\geq 0$ the integral is

$$\text{Pr}(Y-X>z)=\int _z^{\infty }\int _0^{y-z}\frac{n^n n^n \lambda ^n \lambda ^n x^{n-1} y^{n-1} e^{\lambda (-n) x} e^{\lambda (-n) y}}{(n-1)! (n-1)!}dxdy$$

I did not discover a general result but for $n=2, 3, 4, 5$ the results are

$$\frac{1}{2} e^{-2 \lambda z} (\lambda z+1)$$ $$\frac{1}{16} e^{-3 \lambda z} (3 \lambda z (3 \lambda z+5)+8)$$ $$\frac{1}{72} e^{-4 \lambda z} (3 \lambda z (4 \lambda z (4 \lambda z+9)+33)+36)$$ $$\frac{1}{768} e^{-5 \lambda z} (5 \lambda z (5 \lambda z (5 \lambda z (5 \lambda z+14)+87)+279)+384)$$

For $z\leq 0$ the corresponding results are

$$\frac{1}{2} e^{2 \lambda z} (\lambda z-1)+1$$ $$\frac{1}{16} e^{3 \lambda z} (3 \lambda z (5-3 \lambda z)-8)+1$$ $$\frac{1}{24} e^{4 \lambda z} (\lambda z (4 \lambda z (4 \lambda z-9)+33)-12)+1$$ $$1-\frac{1}{768} e^{5 \lambda z} (5 \lambda z (5 \lambda z (5 \lambda z (5 \lambda z-14)+87)-279)+384)$$

Maybe there's a more general answer.

Update

From the answer by @heropup I found that Mathematica has a VarianceGammaDistribution function that can be used to quickly obtain $\text{Pr}(Y-X > z)$ for any specific $n$:

n = 5;
vg = VarianceGammaDistribution[n, n λ, 0, 0];
p = Probability[yMinusx > z, yMinusx \[Distributed] vg]