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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 $P(X_1<X_2)$

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.

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    $\begingroup$ If you fix $X_2$, can you find $P(X_1<X_2)$? $\endgroup$ – SmileyCraft Dec 13 '18 at 14:26
  • $\begingroup$ But won't it include $X_2$ how will I get final result ? $\endgroup$ – Ronald Dec 13 '18 at 14:27
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    $\begingroup$ 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. $\endgroup$ – SmileyCraft Dec 13 '18 at 14:29
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Smiley is right.

Go for finding e.g.:$$P(X_1<X_2)=1-P(X_1\geq X_2)=1-\int_0^{\infty} P(X_1\geq x)f_{X_2}(x)dx=1-\int_0^{\infty} e^{-x}f_{X_2}(x)dx$$

(and leave $X_2-X_1$ for what it is!)

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