Finding probability using joint PDF

Question

Question

Suppose that (jointly continuous) random variables X and Y have joint probability density function

$$f_{X,Y}(x,y) = f(x,y) = e^{-x}, 0<y<x $$

Find $$P(X+Y > 4 | Y = 2)$$

My working $$P(X+Y > 4 | Y = 2) = \frac{P(X+Y > 4 , Y = 2)}{P(Y = 2)}$$

I know that $$\begin{aligned} P(X+Y > 4) & = 1 - P(X+Y < 4)\\ & = 1 - \int_0^2 \int_{y}^{4-y} e^{-x} dxdy \\ & = 1 - \int_0^2 {-e^{-x}} \Big|_y^{4-y} dy \\ & = 1 - \int_0^2 [ - e^{y-4} - (-e^{-y})]dy\\ & = 1 - \left\{[ - e^{y-4} - (e^{-y})]\Big|_0^{2} \right\} \\ & = 2e^{-2} - e^{-4} \end{aligned}$$

Now, I am stuck. Am I correct so far? If so, then how should I find $P(X+Y > 4 , Y = 2)$? I am thinking that this probability should be $0$ because Y is continuous, thus it cannot take on any particular value.

Any intuitive explanations or suggestions will be greatly appreciated!

Answer 1

Notice that $$\mathbb P\{X+Y>4,Y=2\}=\mathbb P\{Y=2\}=0,$$ so it doesn't work.

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You have that $$\mathbb P\{X+Y>4\mid Y=2\}=\int_2^\infty f_{X\mid Y=2}(x)\,\mathrm d x =\int_{2}^\infty \frac{f_{X,Y}(x,2)}{f_Y(2)}\,\mathrm d x,$$

where $$f_Y(y)=\int_0^\infty f_{X,Y}(x,y)\,\mathrm d x.$$

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Edit

$$f_Y(y)=\int_0^\infty f_{X,Y}(x,y)\,\mathrm d x=\int_y^\infty e^{-x}\,\mathrm d x=...$$

Answer 2

Inasmuch as $\,f_{X,Y}(x,y)=0\,$ if $\,x+y<2y\,$ then: $$P[X+Y>2Y]=1,\quad P[X+Y>2Y\mid Y]=1,\quad P[X+Y>4\mid Y=2]=1$$