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Here is the question. The joint PDF of X and Y is given by $f_{XY}(x,y) = {\frac 14} e^{-|x|-|y|}$. Find $P(X \le 1 ,and, Y \le 0)$

Solving the problem I first found the marginal probabilities of X and Y. Can you please explain what I should do next.

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What does $P(X \leq 1, and, Y \leq 0)$ mean? Do you mean the probability of $X \leq 1$ and (at the same time) $Y \leq 0$? Or is it a typo, and do you want $P(X\leq1)$ and $P(Y\leq 0)$ separately? –  fgp May 1 '13 at 7:07
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Probability of both at the same time. –  PasanW May 1 '13 at 7:09

3 Answers 3

up vote 1 down vote accepted

In general you need to use the joint PDF of $(X,Y)$ to calculate such probabilities. To that end, let $A=(-\infty,1]\times(-\infty,0]\subseteq\mathbb{R}^2$, then $$ P(X\leq 1,\, Y\leq 0)=P((X,Y)\in A)=\iint_A f_{(X,Y)}(x,y)\,\mathrm dx\,\mathrm dy\tag{1} $$ which you can easily calculate.

Note that since you have found the marginal PDFs of $X$ and $Y$ you have probably found out that $X$ and $Y$ are independent since $f_{(X,Y)}(x,y)=f_X(x)\cdot f_Y(y)$ for all $x,y$. Hence $$ P(X\leq 1,Y\leq 0)=P(X\leq 1)P(Y\leq 0)=\left(\int_{-\infty}^1f_X(x)\,\mathrm dx\right)\left(\int_{-\infty}^0f_Y(y)\,\mathrm dy\right) $$ which is exactly what you get if you calculate $(1)$.

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To find the probability of some event $E$, you have to compute $$ P(E) = \int_E \:dF $$ where $F$ is your probability distribution. In your case, since you probability distribution has a density, you can also express that as $$ P(E) = \int_E f \:d\lambda $$ where $\lambda$ is the lebesgue measure on your probability space (which then is a subset of some $\mathbb{R}^d)$. The lebesgue measure is the ususal measure or length/area/volume/$\ldots$, so this is just a plain old integral. Additionally, in your case $E = (-\infty,1]\times(-\infty,0]$ is rectangular, which makes the integration especially simple. You get $$ P(E) = \int_E f \:d\lambda = \int_{-\infty}^1\int_{-\infty}^0 f(x,y) \:dx\,dy \text{.} $$

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Solution using the mathStatica / Mathematica combo:

The joint pdf $f(x,y)$ is given as:

f = Exp[-Abs[x] - Abs[y]]/4;     domain[f] = {{x, -∞, ∞}, {y, -∞, ∞}};

You seek:

Prob[x < 1 && y < 0, f]

returns: $\frac 12 - \frac{1}{4 e}$

[ Even if this is homework, I guess your lecturer will want to see some workings, but this will give you something to aim at :) ]

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Which avoids altogether using "the marginal probabilities of X and Y" although the OP requested doing so. // In fact the mathematical insight such a solution can provide per se is remarkably low. To actually do some maths from this, one should at least note that the answer is $\frac12\cdot(1-\frac1{2e})$ and that each of these factors has a precise meaning. But maybe this is what the parenthetical remark at the end uneasily concedes? –  Did May 3 '13 at 11:51

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