P.D.F. with two variables! Help!

**ScienceGeniusGirl** · September 30th 2008, 04:58 PM

I am having some trouble with multiple variables when one is fixed, and vice versa. I was wondering if someone could please help me - I missed a day of class, and it seems like everything was taught while I was sick!

Anyway, if someone could please explain the following problem, I would really appreciate it!

Let the probability density function of variables X and Y equal

f(x,y) = 1/2 when 0<x<y, 0<y<2 and zero elsewhere. Find the conditional means and variances for each variable when another is fixed and the correlation of X and Y.

Thanks in advance!

**mr fantastic** · September 30th 2008, 10:32 PM

Originally Posted by ScienceGeniusGirl

I am having some trouble with multiple variables when one is fixed, and vice versa. I was wondering if someone could please help me - I missed a day of class, and it seems like everything was taught while I was sick!

Anyway, if someone could please explain the following problem, I would really appreciate it!

Let the probability density function of variables X and Y equal

f(x,y) = 1/2 when 0<x<y, 0<y<2 and zero elsewhere. Find the conditional means and variances for each variable when another is fixed and the correlation of X and Y.

Thanks in advance!

Definitions:

Marginal density of Y: $f(y) = \int_{-\infty}^{+\infty} f(x, y) \, dx$ .

Marginal density of X: $f(x) = \int_{-\infty}^{+\infty} f(x, y) \, dy$ .

Conditional density of X given Y = y: $f(x | y) = \frac{f(x, y)}{f(y)}$ for $f(y) > 0$ and 0 otherwise.

Conditional density of Y given X = x: $f(y | x) = \frac{f(x, y)}{f(x)}$ for $f(x) > 0$ and 0 otherwise.

$E(X | Y = y) = \int_{-\infty}^{+\infty} x \, f(x | y) \, dx$ .

$E(Y | X = x) = \int_{-\infty}^{+\infty} y \, f(y | x) \, dy$ .

$Cov(X, Y) = E(XY) - E(X) E(Y)$ $= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x y \, f(x, y) \, dx \, dy - \left( \int_{-\infty}^{+\infty} x \, f(x) \, dx \right) \, \left( \int_{-\infty}^{+\infty} y \, f(y) \, dy \right)$ .

Application of the definitions to the given joint pdf:

Marginal density of Y: $f(y) = \int_{x = 0}^{x = y} \frac{1}{2} \, dx = \frac{y}{2}$ , $x < y < 2$ .

Marginal density of X: $f(x) = \int_{y=x}^{y=2} \frac{1}{2} \, dy = 1 - \frac{x}{2}$ , $0 < x < y$ .

Conditional density of X given Y = y: $f(x | y) = \frac{1/2}{\frac{y}{2}} = \frac{1}{y}$ for $x < y < 2$ .

Conditional density of Y given X = x: $f(y | x) = \frac{1/2}{1 - \frac{x}{2}} = \frac{1}{2-x}$ for $0 < x < y$ .

$E(X | Y = y) = \int_{x=0}^{x=y}\frac{x}{y} \, dx = \frac{y}{2}$ .

$E(Y | X = x) = \int_{y=x}^{y=2} \frac{y}{2-x}\, dy = \frac{2+x}{2}$ .

Cov(X, Y):

$E(XY) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x y \, f(x, y) \, dx \, dy = \int_{y=0}^{y=2} \int_{x=0}^{x=y} \frac{xy}{2} \, dx \, dy$ $= \int_{y=0}^{y=2} \frac{y^3}{4} \, dy = 1$ .

The rest of the calculations are left for you to do.

**ScienceGeniusGirl** · October 2nd 2008, 04:03 PM

Cov(X, Y):

$E(XY) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x y \, f(x, y) \, dx \, dy = \int_{y=0}^{y=2} \int_{x=0}^{x=y} \frac{xy}{2} \, dx \, dy$ $= \int_{y=0}^{y=2} \frac{y^3}{4} \, dy = 1$ .

The rest of the calculations are left for you to do.[/quote]

I guess I don't understand how you got to this point - I understand that if the E(X,Y) is equal to one, then to be calculating the Cov(X,Y), I need to calculate the -E(x)E(y) part - but how does one do that?

I vaguely follow the above posted math, but not strongly enough to calculate the rest of it?

**ScienceGeniusGirl** · October 2nd 2008, 04:13 PM

Mostly, how did you get from the left side of that equation to the right side?

I'm sorry for all the trouble, I just really don't understand.

**mr fantastic** · October 2nd 2008, 07:43 PM

Originally Posted by ScienceGeniusGirl

Mostly, how did you get from the left side of that equation to the right side?

I'm sorry for all the trouble, I just really don't understand.

By definition: $E(XY) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} x y \, f(x, y) \, dx \, dy$

Put in integral terminals that define the region of the xy-plane in which the joint pdf is non-zero: $= \int_{y=0}^{y=2} \int_{x=0}^{x=y} \frac{xy}{2} \, dx \, dy$

To get these terminals, draw the lines y = x and y = 2. 0 < x < y and 0 < y < 2 define the region of the xy-plane that lies between the y-axis and the lines y = x and y = 2. This is the area the integral terminals define. You're expected to have experience with doing this sort of thing.

Integrate with respect to x:

$= \int_{y=0}^{y=2} \left[\frac{x^2 y}{4}\right]_{0}^{y} \, dx \, dy$

(note that y is treated as a constant because the integration is with respect to x)

$= \int_{y=0}^{y=2} \frac{y^3}{4} \, dy$

Integrate with respect to y: $\left[ \frac{y^4}{16}\right]_{0}^{2} = 1$ .

**ScienceGeniusGirl** · October 4th 2008, 03:02 PM

Ok, so I did the calculations for E(x) and E(Y) - Or tried to!

I got E(X) = -1/2
and E(Y) = 1

Therefore, my Cov(X,Y) = 1 - (-1/2)(1) = 1.5 ???

Currently working on my correlation, based on this covariance

**mr fantastic** · October 4th 2008, 09:08 PM

Originally Posted by ScienceGeniusGirl

Ok, so I did the calculations for E(x) and E(Y) - Or tried to!

I got E(X) = -1/2 Mr F says: Since the pdf is non-zero only for 0<x<y, 0<y<2, this cannot possibly be correct. Surely 0 < E(X) < 2 .....!

and E(Y) = 1 Mr F says: Incorrect.

Therefore, my Cov(X,Y) = 1 - (-1/2)(1) = 1.5 ???

Currently working on my correlation, based on this covariance

How did you get those answers?

$E(X) = \int_{-\infty}^{+\infty} x \, f(x) \, dx = \int_{0}^{2} x \, \left(1 - \frac{x}{2} \right) \, dx = \frac{2}{3}$ .

$E(Y) = \int_{-\infty}^{+\infty} y \, f(y) \, dy = \int_{0}^{2} y \, \left(\frac{y}{2} \right) \, dy = \frac{4}{3}$ .

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