0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – Did
    Aug 25 '14 at 12:01
2
$\begingroup$

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: $$\begin{align} \Pr(Y>0\mid X+Y>1) & = \frac{1-2\;e^{-1}}{1-e^{-2}} \\[1ex] & \frac{e^2-2\;e}{e^2-1} \\[1ex] & \approx 0.305{\small 6}\ldots \end{align}$$

$\endgroup$
1
$\begingroup$

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...).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.