A Continuous Multivariate Distribution

[squenshl]

Homework Statement

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 $a = 1$, $b = 2$.

Homework Equations

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 want to make sure that my $c$ is correct first.
Not sure on 3. and 4.

 

[vela]

Homework Statement

The random variable $(x,y)$ has density $f(x,y) = ce^{-(ax+by)}$ for $0\leq y\leq x\leq 1$, with given constants $a > 0$, $b > 0$.

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.

 

[squenshl]

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.

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 $0\leq y\leq x\leq 1$.

 

[vela]

My problem is how to interpret $0\leq y\leq x\leq 1$.

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

 

[Ray Vickson]

Homework Statement

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 $a = 1$, $b = 2$.

Homework Equations

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 want to make sure that my $c$ is correct first.
Not sure on 3. and 4.

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.

 

[squenshl]

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

 

[squenshl]

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

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

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

 

[squenshl]

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

 

[vela]

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

 

[squenshl]

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

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

 

[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
$$\begin{split}
f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\
&= c\int_{0}^{x} e^{-(ax+by)} \; dy?
\end{split}$$

 

[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
$$\begin{split}
f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\
&= c\int_{0}^{x} e^{-(ax+by)} \; dy?
\end{split}$$

Still not sure which it is and why. Please help.

 

[squenshl]

Still not sure which it is and why. Please help.

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

 

[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 want to 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.

 

[squenshl]

I guess I'm correct then.

 

[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 $x$?