# $X_1$ be an exponential random variable with mean $1$ and $X_2$ be a gamma random variable with mean $1$ and variance $2$ find $P(X_1<X_2)$

Let $$X_1$$ be an exponential random variable with mean $$1$$ and $$X_2$$ be a gamma random variable with mean $$2$$ and variance $$2$$. If $$X_1$$ and $$X_2$$ are Independent random variable then $$P(X_1

My input

$$Exp(a)=ae^{-ax} \ \ \ ; G(a,\lambda)=\dfrac{a^{\lambda}}{\Gamma{\lambda}\ }e^{ax}x^{\lambda-1}$$

$$X_1\sim Exp(1)=e^{-x}\implies G(1,1)$$

Mean $$=\dfrac{\lambda}{a}=2 \implies \lambda=2a$$

Variance = $$\dfrac{\lambda}{a^2}=\dfrac{2a}{a^2}=2 \implies a=1,\lambda=2$$

$$X_2\sim G(1,2)$$

$$P(X_2-X_1<0)$$

I am trying to find out distribution of $$X_2-X_1$$ I tried to subtract the parameter($$\lambda_1,\lambda_2$$) but the property of MGF works in addition only. Secondly I tried using Gamma Poisson relationship but my Poisson parameter includes $$X_2$$ so I am out of option I need a hint or something.

• If you fix $X_2$, can you find $P(X_1<X_2)$? – SmileyCraft Dec 13 '18 at 14:26
• But won't it include $X_2$ how will I get final result ? – Ronald Dec 13 '18 at 14:27
• If you have a formula for $P(X_1<X_2)$ for fixed $X_2$, you can then calculate $\int P(X_1<X_2)f_G(X_2)\mbox{d}X_2$ where $f_G$ is the probability density function of the gamma distribution. This method is known as divide and conquer. – SmileyCraft Dec 13 '18 at 14:29

Go for finding e.g.:$$P(X_1
(and leave $$X_2-X_1$$ for what it is!)