# Conditional probability for random variables with different distributions

Random variables $X$ and $Y$ are independent, where $X$ is exponentially distributed with parameter $1$ and $Y$ has uniform distribution on $[-1,1]$ interval. Find $\mathbb{P}(Y>0|X+Y>1)$.

My attempt was to use the definition of conditional probability and calculate $\mathbb{P}(Y>0|X+Y>1)=\frac{\mathbb{P}(Y>0,X+Y>1)}{\mathbb{P}(X+Y>1)}$, but the thing is, I don't know how to find the numerator of this fraction.

$g_X(x)=e^{-x}$

$g_Y(x)=\frac{1}{2}\mathbb{1}_{[-1,1]}(x)$

\begin{align}g_{X+Y}(t) &= \int_{-\infty}^{+\infty}g_X(x)g_Y(t-x)\ dx = \int_{-\infty}^{+\infty}e^{-x}\mathbb{1}_{[-1,1]}(t-x)\ dx = \int_{t-1}^{t+1}e^{-x}\ dx = -e^{-x}\Big |_{t-1}^{t+1} = \\&=\frac{1}{e^{t-1}}-\frac{1}{e^{t+1}} = \frac{e^2-1}{e^{t+1}}\end{align}

$$\mathbb{P}(X+Y>1) = \int_{1}^{\infty}\frac{e^2-1}{e^{t+1}}\ dt = \frac{e^2-1}{e}\int_{1}^{\infty}\frac{1}{e^t} = \frac{e^2-1}{e}\left(-e^{-t}\Big|_1^{\infty}\right) = \frac{e^2-1}{e^2} = 1-\frac{1}{e^2}$$

What I've done so far, is it even correct?

• Did you note that the numerical result proposed in the accepted answer is wrong (hence also some steps of the proof leading to it)? Does it matter?
– Did
Aug 25 '14 at 12:01

This is my way to find the denominator: \begin{align} \Pr(X+Y>1) & = \Pr(Y>1-X) \\[1ex] & = \Pr(Y>1-X, X<2) +\Pr(Y>-1, X\geq 2) \\[1ex] & = \int_0^2 \int_{1-x}^{1} e^{-x}\cdot \frac 12 \operatorname{d}y\operatorname{d}x +\int_2^\infty \int_{-1}^{1} e^{-x}\cdot \frac 12 \operatorname{d}y\operatorname{d}x \\[1ex] & = \frac 12\int_0^2 y\mid_{1-x}^{1} e^{-x} \operatorname{d}x +\frac 12\int_2^\infty y\mid_{-1}^{1} e^{-x} \operatorname{d}x \\[1ex] & = \frac 12\int_0^2 x\; e^{-x} \operatorname{d}x +\frac 12\int_2^\infty 2\; e^{-x} \operatorname{d}x \\[1ex] & = \frac 12\bigl[ (x-1)\; e^{-x}\bigr]_{x=0}^{x=2} + \frac 12\bigl[ 2\; e^{-x}\bigr]_{x=2}^{x\to\infty} \\[1ex] & = \frac 12(e^{-2}+1) + \frac 12(0-2e^{-2}) \\[1ex] & = \frac 12(1-e^{-2}) \end{align} So to find the numerator \begin{align} \Pr(Y>0,X+Y>1) & = \Pr(Y>0,Y>1-X) \\[1ex] & = \Pr(Y>1-X, X<1) +\Pr(Y>0, X\geq 1) \\[1ex] & = \int_0^1 \int_{1-x}^{1} e^{-x}\cdot\frac 12 \operatorname{d}y\operatorname{d}x +\int_1^\infty \int_{-1}^{1} e^{-x}\cdot \frac 12 \operatorname{d}y\operatorname{d}x \\[1ex] & = \frac 12\int_0^1 y\mid_{1-x}^{1} e^{-x} \operatorname{d}x +\frac 12\int_1^\infty y\mid_{-1}^{1} e^{-x} \operatorname{d}x \\[1ex] & = \frac 12\int_0^1 x\; e^{-x} \operatorname{d}x +\frac 12\int_1^\infty 2\; e^{-x} \operatorname{d}x \\[1ex] & = \frac 12\bigl[ (x-1)\; e^{-x}\bigr]_{x=0}^{x=1} + \frac 12\bigl[ 2\; e^{-x}\bigr]_{x=1}^{x\to\infty} \\[1ex] & = \frac 12(0+1) + \frac 12(0-2e^{-1}) \\[1ex] & = \frac 12(1-2e^{-1}) \end{align} Thus: \begin{align} \Pr(Y>0\mid X+Y>1) & = \frac{1-2\;e^{-1}}{1-e^{-2}} \\[1ex] & \frac{e^2-2\;e}{e^2-1} \\[1ex] & \approx 0.305{\small 6}\ldots \end{align}

The usual method, applied as un-cleverly as possible: by independence of $X$ and $Y$, $$P(X+Y\gt1)=\int_0^\infty P(x+Y\gt1)\,f_X(x)\,\mathrm dx=\int_0^\infty P(Y\gt1-x)\,\mathrm e^{-x}\,\mathrm dx.$$ Now, for every $x\gt0$, $P(Y\gt1-x)=\min\left\{\frac12x,1\right\}$ hence $$P(X+Y\gt1)=\int_0^2\tfrac12x\mathrm e^{-x}\,\mathrm dx+\int_2^\infty\mathrm e^{-x}\,\mathrm dx=\ ...$$ Likewise, $$P(Y\gt0,X+Y\gt1)=\int_0^\infty P(Y\gt1-x,Y\gt0)\,\mathrm e^{-x}\,\mathrm dx.$$ Now, for every $x\gt0$, $P(Y\gt1-x,Y\gt0)=\min\left\{\frac12x,\frac12\right\}$ hence $$P(Y\gt0,X+Y\gt1)=\int_0^1\tfrac12x\mathrm e^{-x}\,\mathrm dx+\int_1^\infty\tfrac12\mathrm e^{-x}\,\mathrm dx=\ ...$$ To help you check your computations, the final result is $P(Y\gt0\mid X+Y\gt1)=\frac{\mathrm e}{\mathrm e+1}\approx0.731$ (and the fact that this value is greater than $\frac12$ should come as a relief...).