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$X$ is the exponential distribution $e^{-x}$, $Y$ is the normal distribution with mean $0$ and variance $1$. $X$ and $Y$ are independent. How do I find $P(X<Y+1)$?

My attempt: $$\int_{-1}^{\infty} F_X(y+1) f_Y(y) dy$$ $$=\int_{-1}^{\infty} (1-e^{-(y+1)}) f_Y(y) dy$$ $$=\phi(1)-\int_{-1}^{\infty} (e^{-(y+1)}) f_Y(y) dy$$

Am I on the right track? How do I proceed from here?

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    $\begingroup$ You should mention $X$ and $Y$ are independent. $\endgroup$
    – StubbornAtom
    Commented Feb 17, 2018 at 5:36
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    $\begingroup$ Where does the $\lambda$ come from? You started out correctly. $\endgroup$
    – StubbornAtom
    Commented Feb 17, 2018 at 5:43
  • $\begingroup$ @StubbornAtom My bad there's no lambda. $\endgroup$
    – Code
    Commented Feb 17, 2018 at 5:49
  • $\begingroup$ @Code He just removed it! xD $\endgroup$
    – Maffred
    Commented Feb 17, 2018 at 5:49
  • $\begingroup$ SImply add and remove $3$ to complete the square, then change variable and get again an integral of a normal! $\endgroup$
    – Maffred
    Commented Feb 17, 2018 at 5:53

1 Answer 1

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$$\phi(1)-\int_{-1}^{\infty} (e^{-(y+1)}) f_Y(y) dy $$

$$ = \phi(1)-\frac{1}{\sqrt{2\pi}}\int_{-1}^{\infty} e^{-\frac{1}{2}(y^2+2y+2)}dy$$

$$ = \phi(1)-\frac{e^{-\frac 1 2}}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-\frac{1}{2}t^2} dt $$ where $t=y+1$.

$$ = \phi(1) - \frac{e^{-\frac 1 2}}{2}$$

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