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The joint PDF $f_{X,Y}$: $$f_{X,Y}(x,y) = c e^{-2x} e^{-3y} $$ $$ where \space \space c=15 $$ With bounds: $$ 0 \le x\le y<\infty$$ $$0\text{ otherwise}$$

I want to find multiple probabilities, for example $P[Y > 10]$ and $P[X < 4]$. I know I can find the former by calculating the following equation:

$$\int_{10}^\infty \int_0^y c e^{-2x} e^{-3y}\,dx\,dy $$

However, this would mean I have to work out the expression for every probability. Rather I'd like to find a more general, common, function $F_{X,Y}(x,y)$, from which I can easily extract both probabilities. I tried to work out the following equation but that resulted in an incorrect answer:

$$\int_{0}^y \int_v^x c e^{-2u} e^{-3v}\,du\,dv $$

I am not recognising my own mistakes at this point. Any help and hints would be much appreciated. Thanks.

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    $\begingroup$ Isn't is just the other way round: $\int_{y}^{ \infty}\int_v^x c e^{-2u} e^{-3v}\,du\,dv$ for $P(Y>y)$ and $P(X<x)$? $\endgroup$ Commented Jul 15, 2022 at 19:17
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    $\begingroup$ Or rather $P(Y>y,X\le x) = \int_y^\infty \int_0^{\min(v, x)} ce^{-2u}e^{-3v}dudv$, since $x\le y$? $\endgroup$ Commented Jul 15, 2022 at 19:32
  • $\begingroup$ @callculus42 That would result in $\frac{c}{2}[-\frac{1}{3}e^{-2x}e^{-3y}+\frac{1}{5}e^{-5y}]$, from which $F_Y(y) = F_{X,Y}(\infty, y)$ follows, filling in $y = 10$ then results in a different answer compared to the first mentioned integral. Am I missing something? $\endgroup$
    – Evochrome
    Commented Jul 15, 2022 at 19:37
  • $\begingroup$ @HeiniHøgnason How would I go about working out that expression? $\endgroup$
    – Evochrome
    Commented Jul 15, 2022 at 19:38
  • $\begingroup$ @Evochrome I think Heini Høgnason's comment is right. My term is not true in general. $\endgroup$ Commented Jul 15, 2022 at 19:39

1 Answer 1

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You can compute the probabilities using the joint cumulative distribution function:

$$F_{X,Y}(x, y) = P[X \leq x, Y \leq y].$$

The diagram below shows the region of integration for computing the joint CDF.

Region of integration

Assuming that $x \leq y$ we can compute $F_{X,Y}(x, y)$ as

$$F_{X,Y}(x, y) = \int_{v = 0}^{y} \int_{u = 0}^{\min(x, v)} f_{X,Y}(u, v) \thinspace du \thinspace dv,$$

and if $x > y$ then, due to the restriction that $x < y$ in the support of the PDF, $P[X \leq x, Y \leq y] = P[X \leq y, Y \leq y] = F_{X,Y}(y, y)$.

Now to compute the integral \begin{aligned} F_{X,Y}(x, y) &= \int_{v = 0}^{y} \int_{u = 0}^{\min(x, v)} f_{X,Y}(u, v) \thinspace du \thinspace dv \\ &= c \int_{v = 0}^{y} e^{-3v} \int_{u = 0}^{\min(x, v)} e^{-2u} \thinspace du \thinspace dv \\ &= c \int_{0}^{y} e^{-3v} \left[-\frac{1}{2}e^{-2u}\right]_{u = 0}^{\min(x, v)} \thinspace dv \\ &= c \int_{0}^{y} e^{-3v} \left[\frac{1}{2} - \frac{1}{2}e^{-2\min(x, v)}\right] \thinspace dv \\ &= \frac{c}{2} \int_{0}^{y} e^{-3v} - e^{-3v -2\min(x, v)} \thinspace dv \\ &= \frac{c}{2} \left[\int_{0}^{y} e^{-3v} \thinspace dv - \int_{0}^{x} e^{-3v -2v} \thinspace dv - \int_{x}^{y} e^{-3v - 2x} \thinspace dv\right] \\ &= \frac{c}{2} \left[\left[-\frac{1}{3} e^{-3v}\right]_{0}^{y} - \left[-\frac{1}{5}e^{-5v}\right]_{0}^{x} - e^{-2x}\left[-\frac{1}{3}e^{-3v}\right]_{x}^{y}\right] \\ &= \frac{c}{2} \left[\left[\frac{1}{3} - \frac{1}{3}e^{-3y}\right] - \left[\frac{1}{5} - \frac{1}{5}e^{-5x}\right] - e^{-2x}\left[\frac{1}{3}e^{-3x} - \frac{1}{3}e^{-3y}\right]\right] \\ &= \frac{c}{30} \left[5 - 5e^{-3y} - 3 + 3e^{-5x} - 5e^{-5x} + 5e^{-2x-3y}\right] \\ &= \frac{c}{30} \left[2 - 5e^{-3y} - 2e^{-5x} + 5e^{-2x-3y}\right]. \\ \end{aligned}

We can check a couple of edge cases here: $$F_{X, Y}(0, 0) = \frac{c}{30}\left[2 - 5 - 2 + 5\right] = 0,$$ $$F_{X, Y}(0, \infty)= \frac{c}{30} \left[2 - 0 - 2 + 0\right] = 0, \textrm{and}$$ $$F_{X, Y}(\infty, \infty) = \frac{c}{30}\left[2 - 0 - 0 + 0\right] = 1,$$ since $c = 15$.

You can now use $F_{X, Y}$ to compute any required probabilities. E.g. $$P[Y > 10] = 1 - P[X \leq 10, Y \leq 10] = 1 - F_{X,Y}(10, 10)$$ and $$P[X < 4] = P[X \leq 4, Y \leq \infty] = F_{X, Y}(4, \infty)$$ (here $P[X < 4] = P[X \leq 4]$ as $P[X = 4] = 0$).

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    $\begingroup$ This is exactly what I was hoping to achieve, thanks for explaining it in such a concise and clear manner! $\endgroup$
    – Evochrome
    Commented Jul 17, 2022 at 17:36

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