independent Exponential distribution P(X > Y + 1) $X$ 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 $P(Z>z)=1-P(Z\leq z)$
$1 - P(X-Y\leq z)
=1 - \int_0^\infty\int_0^{y-z}\lambda_1 e^{-\lambda_1x}\lambda_2 e^{-\lambda_2y} dxdy$
 A: 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:
$\begin{align}
\Pr(X-Y> 1) & = \Pr(X > Y+1)
\\
& = \iint_{{\bf X\times Y}: x>y+1} \operatorname{d}^2 F_{X,Y}(x,y)
& \text{by definition}
\\
& = \int_{0}^{\infty} \int_{1+y}^{\infty} \operatorname{d}F_X(x)\operatorname{d}F_Y(y)
& \text{by independence}
\\
& = \int_{0}^{\infty} f_Y(y) \int_{1+y}^{\infty} f_X(x) \operatorname{d}x \operatorname{d}y
& \text{by expansion}
\\
& =\int_{0}^{\infty} f_Y(y) \Pr(X> 1+y)\operatorname{d} y
& \text{by definition}
\\ 
& \color{gray}{= \int_{0}^{\infty} f_Y(y) \Pr(X> 1+Y\mid Y=y) \operatorname{d} y}
& \color{gray}{\text{by independence; just to note}}
\\
& = \int_0^\infty (b\, e^{-by})(e^{-a(1+y)}) \operatorname{d}y
& \text{by substitution from the CDF and pdf}
\\
& = b\,e^{-a}\, \int_0^\infty e^{-(a+b)y}\operatorname{d}y
& \text{by rearranging to simplify}
\\
& = b\,e^{-a}\, \left[ -\frac{ e^{-(a+b)y} }{ a+b } \right]_{y=0}^{y\to\infty}
& \text{by integration}
\\
& = \frac{b\,e^{-a}}{a+b}
& \text{by evaluation}
\\[1ex]
\therefore\quad &\boxed{ \Pr(X-Y> 1)=\dfrac{b\,e^{-a}}{a+b} }
\end{align}$
