# Finding probability using joint PDF

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

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!

• $P(Y=2)=0$ so your first line is not valid. Commented Apr 22, 2023 at 9:36
• You have $f_X(x) = \int_0^x\, e^{-x}\, dy = xe^{-x}$. You're interested in $\operatorname{prob}(X + Y > 4\vert Y=2) = \operatorname{prob}(X > 2) = \int_2^\infty\, f_X(x)\, dx$, which works out to be $3/e^2$. Commented Apr 22, 2023 at 21:01

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

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

Edit

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

• I tried computing the marginal pdf of $Y$ using the integral you suggested but it equates to $1$. It seems strange to me as my answer does not depends on $y$. Am I doing this the right way? Commented Apr 22, 2023 at 11:47
• @Leeeee: I edited my answer.
– Surb
Commented Apr 22, 2023 at 16:59

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

• More simply: $\mathsf P(X+Y> 4\mid Y=2)=\mathsf P(X>2\mid Y=2)=1$ Commented Apr 26, 2023 at 4:26