# Probability of exponential distribution less than normal distribution.

$X$ 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?

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

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