Integration regarding joint probability density functions to find required probabilities

Question

Edit

It seems I could not evaluate $\mathbb{P}(X + Y > 4)$ correctly as I had wrongly assumed that $4 - x$ is always positive, when in actual fact it could be negative, so there was another case to consider, which I had missed. Many thanks to users @MathLover and @StubbornAtom for their answers!

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Question

Suppose that random variables $X$ and $Y$ have joint probability density function (PDF) $$f_{X, Y}(x, y) = e^{-x},\quad 0 < y < x.$$

Find $\mathbb{P}(X + Y > 4)$ and $\mathbb{P}(X + Y > 4\ |\ Y = 2)$.

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My working

\begin{align} \mathbb{P}(X + Y > 4) & = \mathbb{P}(Y > 4 - X) \\[5 mm] & = \int^{\infty}_0 \int^x_{4 - x} e^{-x}\ \mathrm{d}y\ \mathrm{d}x \\[5 mm] & = -2 \end{align}

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The answer clearly cannot be $-2$, since we are dealing with probability and my best guess is my limits of integration are incorrect. In particular, I think the support for the given joint PDF is throwing me off - I am not sure how to interpret it.

I also have no clue how to proceed with $\mathbb{P}(X + Y > 4\ |\ Y = 2)$, so any hints for this part will be nice as well :)

Answer 1

You missed that $0 \lt y \lt x$. At intersection of $y = 4-x$ and $y = x, x = 2$. Also at intersection of $y = 0$ and $y = 4 - x, x = 4$

$\displaystyle \mathbb{P}(X + Y > 4) = \int_2^4 \int^x_{4 - x} e^{-x}\ \mathrm{d}y\ \mathrm{d}x + \int_4^{\infty} \int_0^x e^{-x}\ \mathrm{d}y\ \mathrm{d}x$

Answer 2

The bounds of integration are incorrect, since you have to incorporate the fact that support of $(X,Y)$ is the set $S=\{(x,y)\in \mathbb R^2:0<y<x\}$. This means joint density of $(X,Y)$ is actually $f_{X,Y}(x,y)=e^{-x} \mathbf1_{0<y<x}$ where $\mathbf1$ is an indicator function. In other words, $f_{X,Y}(x,y)=e^{-x}$ if $(x,y)\in S$ and $f_{X,Y}(x,y)=0$ if $(x,y)\notin S$.

Therefore,

\begin{align} P(Y>4-X)&=\int_0^\infty\left\{\int_{\max(0,4-x)}^x e^{-x}\,\mathrm dy\right\}\mathbf1_{x>4-x}\,\mathrm dx \\&=\int_2^4 e^{-x}\int_{4-x}^x \,\mathrm dy\,\mathrm dx+\int_4^\infty e^{-x} \int_0^x \,\mathrm dy\,\mathrm dx \\&=\cdots \end{align}

The other probability is $P(X>2\mid Y=2)$, for which you require the conditional density of $X$ given $Y$. Now the joint density factors as

$$f_{X,Y}(x,y)=e^{-(x-y)}\mathbf1_{x>y} \cdot e^{-y}\mathbf1_{y>0}$$

This suggests that the conditional density of $X$ given $Y$ is

$$f_{X\mid Y}(x\mid y)=e^{-(x-y)}\mathbf1_{x>y}$$

Hence, $$P(X>2\mid Y=2)=\int_2^\infty f_{X\mid Y}(x\mid 2)\,\mathrm dx=\cdots$$