Homework Help: A Continuous Multivariate Distribution

1. Apr 9, 2016

squenshl

1. The problem statement, all variables and given/known data
The random variable $(x,y)$ has density $f(x,y) = ce^{-(ax+by)}$ for $0\leq y\leq x\leq 1$, with given constants $a > 0$, $b > 0$.
1. Compute the constant $c$.
2. Find the conditional probability density $f_y(y|x)$.
3. Compute the regression curve of $Y$ on $X$ i.e. $E(Y|X = x)$.
4. Sketch this regression curve for $a = 1$, $b = 2$.

2. Relevant equations

3. The attempt at a solution
For 1. my limits of integration for both $x$ and $y$ is $0$ and $1$. This gives (after doing the double integration) $c = \frac{ab}{e^{-(a+b)}-e^{-a}-e^{-b}+1}$. Is this right???
For 2. the conditional density is given by the joint density divided by the marginal density of $x$. I just wanna make sure that my $c$ is correct first.
Not sure on 3. and 4.

2. Apr 10, 2016

vela

Staff Emeritus
Did you write the conditions on $x$ and $y$ exactly as they were given to you? If so, the limits on your integral and your answer to #1 are wrong.

3. Apr 10, 2016

squenshl

I did $c\int_0^1 \int_0^1 e^{-(ax+by)} \; dx \; dy = 1$ and solved for $c$. If this is wrong, then my problem is how to interpret $0\leq y\leq x\leq 1$.

4. Apr 10, 2016

vela

Staff Emeritus
You need to take into account that it says $y \le x$.

5. Apr 10, 2016

Ray Vickson

In problems involving two-dimensional densities, always sketch the correct (x,y)-region before you start trying to do integration. Your error will stare you in the face if you do that.

6. Apr 10, 2016

squenshl

Oh ok so in that case I get $y$ from $0$ to $x$ and $x$ from $0$ to $1$.

7. Apr 10, 2016

squenshl

Using these new limits i get $c = \frac{ab(a+b)}{e^{-(a+b)}-(a+b)e^{-a}+b}$

8. Apr 10, 2016

squenshl

Using these new limits i get $c = \frac{ab(a+b)}{e^{-(a+b)}-(a+b)e^{-a}+b}$

9. Apr 10, 2016

vela

Staff Emeritus
It looks like you made an algebra mistake somewhere. You should have gotten
$$c = \frac{ab(a+b)}{a e^{-(a+b)}-(a+b)e^{-a}+b}.$$

10. Apr 10, 2016

squenshl

Yes I did get that just forgot to put the $a$ in there haha. Thanks.

11. Apr 10, 2016

squenshl

To find the marginal density do I calculate
$$\begin{split} f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\ &= c\int_{0}^{1} e^{-(ax+by)} \; dy??? \end{split}$$
or is it
$$\begin{split} f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\ &= c\int_{0}^{x} e^{-(ax+by)} \; dy??? \end{split}$$

12. Apr 11, 2016

squenshl

13. Apr 11, 2016

squenshl

Never mind. Pretty sure it's the second one.

14. Apr 12, 2016

squenshl

I'm asked to find the conditional probability function $f_y(y|x)$. This is
$$\begin{split} f_y(y|x) &= \frac{f(x,y)}{f_x(x)} \\ &= \frac{ce^{-(ax+by)}}{\frac{c}{b}\left[e^{-ax}-e^{-(ax+bx)}\right]} \\ &= \frac{be^{-(ax+by)}}{e^{-ax}-e^{-(ax+bx)}}. \end{split}$$
Just wanna make sure I'm good.
I'm also asked to find the regression curve (which is the one I'm struggling with) $E(Y|X=x)$. I get
$$\begin{split} E(Y|X=x) &= \int_0^1 y f_y(y|x) \; dy \\ &= \int_0^1 \frac{bye^{-(ax+by)}}{e^{-ax}-e^{-(ax+bx)}} \\ &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \int_0^1 ye^{-by} \; dy. \end{split}$$
We use integration by parts on the integral above. Let $u=y$, so $du=dy$. Also $dv = e^{-by}$, so $v = -\frac{e^{-by}}{b}$. Hence,
$$\begin{split} E(Y|X=x) &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \int_0^1 ye^{-by} \; dy \\ &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \left(\left[-\frac{ye^{-by}}{b}\right]_0^1+\frac{1}{b}\int_0^1 e^{-by} \; dy\right) \\ &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \left(\left[-\frac{ye^{-by}}{b}\right]_0^1-\left[\frac{e^{-by}}{b^2}\right]_0^1\right) \\ &= \frac{e^{-ax}}{e^{-ax}-e^{-(ax+bx)}}\frac{1-(b+1)e^{-b}}{b^2}. \end{split}$$
My question is are the limits of integration $0$ to $1$ shown or $0$ to $x$ or even another set of limits???
Thanks again.

15. Apr 13, 2016

squenshl

I guess I'm correct then.

16. Apr 14, 2016

squenshl

Nope that is wrong.
$$E(Y|X=x) = \int_0^x \frac{bye^{b(x-y)}}{e^{bx}-1} \; dy = \frac{1}{b}-\frac{x}{e^{bx}-1}$$
after several lines of working.
Is this right as my limits of integration are $0$ and $x$???