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$X$ and $Y$ are independent exponentially distributed random variables with parameters $a$ and $b$.

Calculate $P(X > Y + 1)$.

I have let $X-Y=Z$ and Then $P(Z>z)=1-P(Z\leq z)$

$1 - P(X-Y\leq z) =1 - \int_0^\infty\int_0^{y-z}\lambda_1 e^{-\lambda_1x}\lambda_2 e^{-\lambda_2y} dxdy$

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    $\begingroup$ Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? $\endgroup$ – 5xum Jul 29 '14 at 8:44
  • $\begingroup$ I was going to let $X-Y=Z$ then calculate the $P(Z>z)$. $F_Z(z)=P(Z>z)$ $\endgroup$ – rmvdwalt Jul 29 '14 at 8:58
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    $\begingroup$ That sounds like a good idea. Try it out and edit your question with the results you get. $\endgroup$ – 5xum Jul 29 '14 at 9:02
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    $\begingroup$ Why does $P(Z>z).F_Z(z)$ equal $P(Z>z)$ for all $z$? Are you assuming that $F_Z(z) = 1$ for all $z$? Hint: If you do want to use conditional probabilities, then try finding $P(X > Y+1\mid Y = y)$ which should have the same value as $P(X > y+1)$ (why?) and then find $$P(X>Y+1) = \int_{-\infty}^\infty P(X>Y+1\mid Y = y)f_Y(y)\,\mathrm dy$$ which should be a lot easier than finding $F_Z(z)$ as in your idea. $\endgroup$ – Dilip Sarwate Jul 29 '14 at 9:21
  • $\begingroup$ Can it be done without conditional probabilities $\endgroup$ – rmvdwalt Jul 29 '14 at 9:25
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We know the distributions, and hence CDF and pdf of X and Y. Further we know that the random variables are independent.

$$\begin{align}X\bot Y & \iff f_{X,Y}(x,y)=f_X(x)f_Y(y), \forall (x,y)\in {\bf X\times Y} \\[1ex] X\sim{\cal Exp}(a) & \iff \Pr(X\leq x)= (1-e^{-ax})\operatorname{\bf 1}_{[0, \infty)}(x) \\ & \iff f_X(x)= a\,e^{-ax}\operatorname{\bf 1}_{[0,\infty)}(x) \\[1ex] Y\sim{\cal Exp}(b) & \iff \Pr(Y\leq y)= (1-e^{-by})\operatorname{\bf 1}_{[0, \infty)}(y) \\ & \iff f_Y(y)= b\,e^{-by}\operatorname{\bf 1}_{[0,\infty)}(y) \end{align}$$

Thus we can evaluate using a double integral, or by conditional and total probability. The later is possible because we already know what the integral of the probability distribution function will be.

However, for completeness:

$\begin{align} \Pr(X-Y> 1) & = \Pr(X > Y+1) \\ & = \iint_{{\bf X\times Y}: x>y+1} \operatorname{d}^2 F_{X,Y}(x,y) & \text{by definition} \\ & = \int_{0}^{\infty} \int_{1+y}^{\infty} \operatorname{d}F_X(x)\operatorname{d}F_Y(y) & \text{by independence} \\ & = \int_{0}^{\infty} f_Y(y) \int_{1+y}^{\infty} f_X(x) \operatorname{d}x \operatorname{d}y & \text{by expansion} \\ & =\int_{0}^{\infty} f_Y(y) \Pr(X> 1+y)\operatorname{d} y & \text{by definition} \\ & \color{gray}{= \int_{0}^{\infty} f_Y(y) \Pr(X> 1+Y\mid Y=y) \operatorname{d} y} & \color{gray}{\text{by independence; just to note}} \\ & = \int_0^\infty (b\, e^{-by})(e^{-a(1+y)}) \operatorname{d}y & \text{by substitution from the CDF and pdf} \\ & = b\,e^{-a}\, \int_0^\infty e^{-(a+b)y}\operatorname{d}y & \text{by rearranging to simplify} \\ & = b\,e^{-a}\, \left[ -\frac{ e^{-(a+b)y} }{ a+b } \right]_{y=0}^{y\to\infty} & \text{by integration} \\ & = \frac{b\,e^{-a}}{a+b} & \text{by evaluation} \\[1ex] \therefore\quad &\boxed{ \Pr(X-Y> 1)=\dfrac{b\,e^{-a}}{a+b} } \end{align}$

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