Jointly distributed random variables, finding a conditional probability involving a sum.

Question

$X$ and $Y$ are jointly continuous random variables with joint density $f(x, y) = \frac{4}{3}(x+y)e^{-y-2x}$ for $0 \leq x, y < \infty$. Find $\Pr\left(X > k \mid X+Y < k+1\right)$.

My thoughts:

Condition the desired probability on $Y$:

$$ \Pr\left(X > k \mid X+Y < k+1\right) = \int_{y}\Pr\left(X > k \mid X < k+1 - y\right) \Pr\left(Y=y\right) $$

Then the two things that we need to find are $\Pr\left(X > k | X < k+1 - y\right)$ and $\Pr\left(Y=y\right)$. The latter is simply the density of $Y$, which can be found by integrating the joint density over $x$:

$$ f_Y(y) = \int_{0}^{\infty} f(x, y)\,dx $$

Then, we need to find $\Pr\left(X > k \mid X < k+1 - y\right)$, which I think I can find by: $$\Pr\left(X > k \mid X < k+1 - y\right) = \int_{k < X < (k+1-y)}f_X(x)\,dx = \int_{k}^{(k+1-y)}f_X(x)\,dx$$

I am fairly certain that these integrals can be computed by hand with integration by parts, but this seems like a very laborious way of solving the initial probability, and I am wondering if there is an alternative way of thinking about this problem? Or is there a flaw in the procedure I have? Thanks!

Answer 1

My advice: Take an $xy$ plane and draw $y=k+1-x$ with the assumption $k>0$. You need $X\geq 0$ and $Y\geq 0$ so restrict to the first quadrant. The region $X+Y<k+1$ is the same as $Y<k+1-X$ so this region is the right triangle.

Now draw the vertical line $x=k$. Within this space, you want to calculate the probability over the small triangle and divide by the probability of the whole triangle.

Answer 2

Condition the desired probability on $Y$:

$$ \Pr\left(X > k \mid X+Y < k+1\right) = \int_{y}\Pr\left(X > k \mid X < k+1 - y\right) f_Y(y)\,\mathrm d y $$

No. $X$ and $Y$ are not independent. They have a joint distribution which is not seperable into a product of monovariate distributions. So you would need to find

$$ \Pr\left(X > k \mid X+Y < k+1\right) = \int_{}\Pr\left(X > k \mid X < k+1 - y, Y=y\right)f_Y(y)\,\mathrm d y $$

Which is more effort than it is worth.

Rather, just use the normal definition for conditional probability (noting that surely $0\leqslant Y$ and $0\leqslant X$):

$$\begin{align}\mathsf P(X>k\mid X+Y<k+1)&=\dfrac{\mathsf P(X+Y<k+1\cap k<X)}{\mathsf P(X+Y<k+1)}\\[1ex] &=\dfrac{\mathsf P((0\leqslant Y<1)\cap (k<X<k+1-Y))}{\mathsf P((0\leqslant Y<k+1)\cap (0\leqslant X< k+1-Y))}\\[1ex]&=\dfrac{\int_0^1\int_k^{k+1-y} f_{\small X,Y}(x,y)~\mathrm d x~\mathrm d y}{\int_0^{k+1}\int_0^{k+1-y}f_{\small X,Y}(x,y)~\mathrm d x~\mathrm d y}\end{align}$$