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A project has two phases. Let $X$ be the time (in months) required to do phase 1 and Y be the total time (in months) it takes to complete both phases. Suppose that the joint density of $X$ and $Y$ is $f(x,y) = c$, where $0 \leq x^2 \leq y \leq 1$, and otherwise $f(x,y) = 0$.

I found the value of the constant $c$ to be $\frac{3}{2}$, and now I'm trying to use this to find the marginal density of $X$ and the conditional probability that the total time is at least 3/4 of a month given that phase 1 takes 1/2 a month. But I'm stumped.

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Are you sure about $x^2 \le y$? It looks strange –  leonbloy May 10 '13 at 17:28
    
Sketch a diagram of the $x$-$y$ plane and indicate on it the region where the pdf is nonzero. To find $f_X(\cdot)$ at a fixed point $x_0$, you need to compute $\int_{-\infty}^\infty f(x_0,y) dy$ where the integral runs along the line $x=x_0$ in the plane. Draw several such lines for different values of $x_0$ and figure out the value of the corresponding integral. See if you can figure out a pattern to fill in the blanks below: $$f_X(x)=\begin{cases} \quad & \quad \leq x \leq \quad,\\0,&\text{otherwise.}\end{cases}$$ –  Dilip Sarwate May 10 '13 at 17:36
    
The region does not seem consistent with the problem statement. Y (total time) can never be less than X (time of phase 1). –  leonbloy May 10 '13 at 17:50
    
It is possible that the problem is as described, but the region described is not consistent with the natural condition that $Y\ge X$. –  André Nicolas May 10 '13 at 18:54

2 Answers 2

up vote 0 down vote accepted

Your value for $c$ is correct. As for the marginal of $X$, you have everything you need, you just need to integrate the joint density over all possible values of $Y$. That is,

$$f_X(x) = \int_{-\infty}^\infty f_{X,Y}(x,y) \, dy.$$

It's harder to see why this is true for continuous random variables because the density functions don't actually give you probabilities. But, if you imagine that we're in discrete world for a second, then this is the equivalent to taking the sum of $P(X = x, Y = y)$ over all possible values of $Y$, and so what you're left with is just the distribution of $X$. In this case you're just integrating out the $y$'s instead of summing.

The trickiest part can be finding the right limits of integration. Since you're dealing with time, $x$ and $y$ both need to be positive (and satisfy the inequalities given in the definition of the joint pdf).

Finally, once you integrate and find $f_X(x)$ you can solve the next part of the problem. For that you'll want to find the conditional density of $Y$ given $X = x$, which is

$$f_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_X(x)}.$$

In your case you will actually want the conditional distribution of $Y$ given $X = 1/2$, i.e.

$$f_{Y|X}(y\,|\,1/2) = \frac{f_{X,Y}(1/2\, ,\,y)}{f_X(1/2)}.$$

Then using this distribution you can find $P(Y \geq 3/4 \mid X = 1/2)$ by integrating $f_{Y|X}(y\mid \frac{1}{2})$ over the appropriate region.

I don't want to do all these computations for you (since it's a hw problem), but hopefully this gives you enough to get the ball rolling.

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Thank you! Does this look good: $f_X(x) = \int_{y=x^2}^{y=1} \frac{3}{2} \, dy = \frac{3}{2}(1-x^2)$? And then $f_{Y|X}(y\,|\,1/2) = \frac{3/2}{\frac{3}{2}(1-(1/2)^2)} = \frac{3/2}{9/8} = 4/3$. Then I can just integrate this from $y=3/4$ to $y=1$, right? –  jimmy May 10 '13 at 17:43
    
@jimmy Nice work!! –  sol May 10 '13 at 17:45

The marginal can be computed by integrating $y$ out.

$f_X(x) = \int f(x,y) dy = \int_{x^2 \le y \le 1} c dy = c(1-x^2), 0 \le x \le 1 $

The conditional probability can be found from the conditional pdf

$f(y|x) = \frac{f(x,y)}{f_X(x)}$

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