Can Independence Simplify Calculating Expectation Values in Probability?

[KayDee01]

Homework Statement

$f(x,y)=6a^{-5}xy^{2}$ $0≤x≤a$ and $0≤y≤a$, $0$ elsewhere

Show that $\overline{xy}=\overline{x}.\overline{y}$

Homework Equations

$\overline{x}=\int^{∞}_{-∞}{x.f(x)dx}$

The Attempt at a Solution

$\overline{x}=\int^{∞}_{-∞}{x.f(x)dx}$
$=\int^{a}_{0}{x.6a^{-5}xy^{2}dx}$
$=6a^{-5}\int^{a}_{0}{x^{2}y^{2}dx}$
$=6a^{-5}$$\frac{1}{3}a^{3}y^{2}$
$=2a^{-2}y^{2}$

Following the same process I get $\overline{y}=\frac{3}{2}a^{-1}x^{2}$

But when it comes to $\overline{xy}$ I'm not really sure how to approach it

 

[LeonhardEu]

You found correcly x¯. Do the same for y and xy and complete your proof. What is your problem?

 

[KayDee01]

You found correcly x¯. Do the same for y and xy and complete your proof. What is your problem?

Sorry, I submitted the question to early and had to add the rest of the problem. I'm not sure what the $\overline{xy}$ integral should look like.

 

[KayDee01]

I tried $\overline{xy}=\int^{∞}_{-∞}{x.y.f(x,y)dx}$
$=\int^{a}_{0}\int^{a}_{0}{x.y.6a^{-5}xy^{2}dx}$
$=\int^{a}_{0}\int^{a}_{0}{6a^{-5}x^{2}y^{3}dx}$
$=\frac{1}{2}a^{2}=/=\overline{x}.\overline{y}$

 

[KayDee01]

I tried

$\overline{xy}=\int^{∞}_{-∞}{x.y.f(x,y)dx}$
$=\int^{a}_{0}\int^{a}_{0}{x.y.6a^{-5}xy^{2}dx}$
$=\int^{a}_{0}\int^{a}_{0}{6a^{-5}x^{2}y^{3}dx}$
$=\frac{1}{2}a^{2}$
which does not equal $\overline{x}.\overline{y}$

 

[LeonhardEu]

Your y¯ has a minor arithmetical mistake but you can find it. And for your major problem: d(xy) = ydx + xdy. Tell me when you have this ;)

Edit: It's not a double integral. It is one dimensional.

 

[KayDee01]

I see me mistake with $\overline{y}$, the x shouldn't be squared right?

If d(x,y)=xdy+ydx. Surely I end up with $\int{xy.f(x,y)d(xy)}=\int{x^{2}y.f(x,y)dy}+\int{xy^{2}.f(x,y)dx}$
And I still can't get that to equal $\overline{x}.\overline{y}$

 

[statdad]

This is an expected value problem? Try evaluating
$$\int_0^a \int_0^a xy f(x,y) \, dx dy$$

again. This
"And for your major problem: d(xy) = ydx + xdy."
has nothing to do with the problem.

 

[HallsofIvy]

The first thing you need to do is define your terms! Your "$\overline{x}$" is "the mean value of x for fixed y" and for "$\overline{y}$" is "the mean value of y for fixed x". Since the first is a function of y and the second a function of x, their product can't possibly be equal to $\overline{xy}$ which is a number.

 

[KayDee01]

This is an expected value problem? Try evaluating
$$\int_0^a \int_0^a xy f(x,y) \, dx dy$$

again. This
"And for your major problem: d(xy) = ydx + xdy."
has nothing to do with the problem.

I did this integral and got $\frac{1}{2}a^{2}$ which doesn't equal $\overline{x}.\overline{y}$ as I am trying to prove. The double integral will always eliminate the variables x and y from the equation. But $\overline{x}.\overline{y}$ has the variables in it.

How do I overcome this?

 

[CAF123]

I did this integral and got $\frac{1}{2}a^{2}$ which doesn't equal $\overline{x}.\overline{y}$ as I am trying to prove. The double integral will always eliminate the variables x and y from the equation. But $\overline{x}.\overline{y}$ has the variables in it.

How do I overcome this?

Your $\overline{xy}$ is correct, so your error is in evaluating $\overline{x}$ and $\overline{y}$. An expectation calculation should return a number.

 

[HallsofIvy]

Yes, which means that either your definitions (of $\overline{x}$, $\overline{y}$, and/or $\overline{xy}$ are wrong or $\overline{x}\overline{y}$ is NOT equal to $\overline{xy}$.

You might want to consider the possibility that the correct definition of $\overline{u}$ for any function u, of x and y, is $\int\int u f(x,y)dydx$ so that, in particular, $\overline{x}= \int_0^a\int_0^a x f(x,y)dydx$, NOT "$\int_0^a xf(x,y)dx$".

(You also posted this under "precalculus homework". Do not do that. Double posting can get you banned.)

 

[KayDee01]

Yes, which means that either your definitions (of $\overline{x}$, $\overline{y}$, and/or $\overline{xy}$ are wrong or $\overline{x}\overline{y}$ is NOT equal to $\overline{xy}$.

You might want to consider the possibility that the correct definition of $\overline{u}$ for any function u, of x and y, is $\int\int u f(x,y)dydx$ so that, in particular, $\overline{x}= \int_0^a\int_0^a x f(x,y)dydx$, NOT "$\int_0^a xf(x,y)dx$".

(You also posted this under "precalculus homework". Do not do that. Double posting can get you banned.)

Using that definition of $\overline{u}$ I got them to equal one another :) Problem solved!
And thanks for the warning about double posting, I won't be doing that again.

 

[jtbell]

The two threads have been merged. As HallsOfIvy noted, please post a question only once. If we think it really belongs in a different forum, we'll move it.

 

[Ray Vickson]

Homework Statement

$f(x,y)=6a^{-5}xy^{2}$ $0≤x≤a$ and $0≤y≤a$, $0$ elsewhere

Show that $\overline{xy}=\overline{x}.\overline{y}$

Homework Equations

$\overline{x}=\int^{∞}_{-∞}{x.f(x)dx}$

The Attempt at a Solution

$\overline{x}=\int^{∞}_{-∞}{x.f(x)dx}$
$=\int^{a}_{0}{x.6a^{-5}xy^{2}dx}$
$=6a^{-5}\int^{a}_{0}{x^{2}y^{2}dx}$
$=6a^{-5}$$\frac{1}{3}a^{3}y^{2}$
$=2a^{-2}y^{2}$

Following the same process I get $\overline{y}=\frac{3}{2}a^{-1}x^{2}$

But when it comes to $\overline{xy}$ I'm not really sure how to approach it

In addition to what others have said: I don't know if you have yet met with the concept of independence, but this problem fits that profile.

You can write your bivariate density f(x,y) as a product of two univariate densities g(x) and h(y):
$$f(x,y) = 6a^{-5}xy^{2} = (2 a^{-2} x) ( 3 a^{-3} y^2) = g(x) h(y), 0 \leq x,y \leq a.$$
Here the individual factors $g(x) = 2 a^{-2} x$ and $h(y) = 2 a^{-3} y^2$ are both univariate probability densities of random variables $X, Y$ on $[0,a]$: they are ≥ 0 and integrate to 1. There is a general theorem that $E(XY) = EX \cdot EY$ if $X$ and $Y$ are independent. You have shown one special case of this. Note: $EX$ is the expectation of $X$, and is the same as what you call $\bar{X}$.