# Probability $X_1 + X_2+ X_3 \geq 1$

Let $X_1,X_2,X_3$ be independent random variables, each of which has the exponential distribution with parameter $\lambda=2$. Estimate the probability that $X_1+X_2+X_3 \geq1$.

I have tried to integrate $2e^{-2x}$ from $0$ to $1$ and subtract this value from $1$, but it was wrong.

I've also tried $e^{-\frac{1}{2}}*(1+\frac{1}{2}+\frac{1}{8})$, but this was wrong too.

Any help would be greatly appreciated.

• Did u know that summing random variables convolves their distributions? – enthdegree Mar 21 '14 at 5:01
• No, so what does that mean? – Liliana Mar 21 '14 at 5:05

## 3 Answers

HINT.
Since $X_i$ has an exponential distribution then $Y=\sum_{i=1}^3\sim \mathrm{Erlang}(3,\, \lambda)\,$ has an Erlang distribution with the probability density function $$f(x; k,\mu)=\frac{ x^{k-1} e^{-\frac{x}{\mu}} }{\mu^k \Gamma(k)}\quad\mbox{for }x, \mu \geq 0$$ where $\mu=\frac{1}{\lambda}$.

The sum of random variables has a distribution that equals the convolution of all their distributions. Fortunately, all your distributions are easy to convolve with each other. The distribution of $X=X_1+X_2+X_3$ is $(2e^{-2x}\cdot u(x))\ast(2e^{-2x}\cdot u(x))\ast(2e^{-2x}\cdot u(x))$ ($\ast$ denoting convolution, u(x) denoting the step function: $1$ when $x\geq 0$ and $0$ otherwise).

Convolution $f\ast g$ is defined as $\int_{-\infty}^{\infty}f(t)g(x-t)dt$. Let's evaluate the first convolution:

$$\int_{-\infty}^{\infty}2e^{-2t}\cdot u(t)\cdot 2e^{-2(x-t)}\cdot u(x-t)dt=\int_{0}^{\infty}4e^{-2x}\cdot u(x-t)dt=4 xe^{-2x}u(x)$$

It remains to convolve $4xe^{-2x}$ with $2e^{-2x}$. I'll let you work out the details for the second convolution (they're largely the same), but it comes out to be:

$$P_X(x)=4x^2e^{-2x}u(x)$$

It remains to integrate this from $1$ to $\infty$:

$$\int_{1}^{\infty}4x^2e^{-2x}u(x)dx=\int_{1}^{\infty}4x^2e^{-2x}dx=\dots=\ell_{n\rightarrow\infty}\left.\left[-2x^2e^{-2x}-2xe^{-2x}-2e^{-2x}\right]\right|_{x=1}^{x=n}$$

$$=0-(-5e^{-2})=5e^{-2}\approx 0.68$$

Why not use the definition and only first principles? For every Borel set $B$, the independence of $(X_1,X_2,X_3)$ yields $$P(X_1+X_2+X_3\in B)=\iint\!\!\!\int_{x_1+x_2+x_3\in B} f_{X_1}(x_1)f_{X_2}(x_2)f_{X_3}(x_3)\mathrm dx_1\mathrm dx_2\mathrm dx_3.$$ Thus, $$P(X_1+X_2+X_3\geqslant1)=8\iint\!\!\!\int_{x_1+x_2+x_3\geqslant1}\mathrm e^{-2(x_1+x_2+x_3)}\mathrm dx_1\mathrm dx_2\mathrm dx_3.$$ An easier version might be $$P(X_1+X_2+X_3\leqslant1)=8\int_0^1\mathrm e^{-2x_3}\int_0^{1-x_ 3}\mathrm e^{-2x_2}\int_0^{1-x_3-x_2}\mathrm e^{-2x_1}\mathrm dx_1\mathrm dx_2\mathrm dx_3.$$ Now perform the $x_1$-integration, then the $x_2$-integration, and finally the $x_3$-integration.