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$X$ 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!

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    $\begingroup$ After trying to perform the integrations, I got a divergent answer at the end, so I think this is the wrong approach to the question, but I am not sure what a correct approach could be? $\endgroup$
    – Eoin S
    Apr 2, 2020 at 19:47

2 Answers 2

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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.

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  • $\begingroup$ Ah, I see, so setting it up that way, the probability of each region would simply be the integral of the joint density over that specific region? For example, let the big right triangle be region A, the trapezoidal region be B, and the small right triangle C. The final probability would then be $\frac{\iint_{C}{f(x,y)dxdy}}{\iint_{A=B+C}{f(x,y)dxdy}}$? $\endgroup$
    – Eoin S
    Apr 2, 2020 at 21:45
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    $\begingroup$ This is exactly how I would do it. I slightly miss spoke in my answer when I called it a trapezoid; I corrected it. But you are correct. $\endgroup$
    – ProfOak
    Apr 3, 2020 at 19:17
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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 $0\leqslant X$):

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

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  • $\begingroup$ Would this be similar to the method described in the other answer by thinking about the problem graphically? There would be a large right triangle, call it Region A, in the first quadrant defined by the line y = (k+1) - x. Then, a vertical line at x = k would divide Region A into Regions B (a trapezoid) and Region C (a smaller right triangle). Then the final probability we are looking for is $\frac{\iint_{C}{f_{X,Y}(x,y)dxdy}}{\iint_{A}{f_{X,Y}(x,y)dxdy}}$ $\endgroup$
    – Eoin S
    Apr 3, 2020 at 19:09
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    $\begingroup$ Yes. $$\begin{align}C&=\{\langle x,y\rangle: 0\leq y<1, k<x<k+1-y\}\\&=\triangle\langle k,0\rangle\langle k,1\rangle\langle k+1,0\rangle\\[2ex]A&=\{\langle x,y\rangle: 0\leq y<k+1, 0<x<k+1-y\}\\&=\triangle\langle 0,0\rangle\langle 0,k+1\rangle\langle k+1,0\rangle\end{align}$$. $\endgroup$ Apr 3, 2020 at 21:55

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