Can Independence Simplify Calculating Expectation Values in Probability?

[KayDee01]

Homework Statement

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

Show that [itex]\overline{xy}=\overline{x}.\overline{y}[/itex]

Homework Equations

[itex]\overline{x}=\int^{∞}_{-∞}{x.f(x)dx}[/itex]

The Attempt at a Solution

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

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

But when it comes to [itex]\overline{xy}[/itex] 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 [itex]\overline{xy}[/itex] integral should look like.

 

[KayDee01]

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

 

[KayDee01]

I tried

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

 

[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 [itex]\overline{y}[/itex], the x shouldn't be squared right?

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

 

[statdad]

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

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 "[itex]\overline{x}[/itex]" is "the mean value of x for fixed y" and for "[itex]\overline{y}[/itex]" 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 [itex]\overline{xy}[/itex] which is a number.

 

[KayDee01]

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

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

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

How do I overcome this?

 

[CAF123]

I did this integral and got [itex]\frac{1}{2}a^{2}[/itex] which doesn't equal [itex]\overline{x}.\overline{y}[/itex] as I am trying to prove. The double integral will always eliminate the variables x and y from the equation. But [itex]\overline{x}.\overline{y}[/itex] 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 [itex]\overline{x}[/itex], [itex]\overline{y}[/itex], and/or [itex]\overline{xy}[/itex] are wrong or [itex]\overline{x}\overline{y}[/itex] is NOT equal to [itex]\overline{xy}[/itex].

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

(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 [itex]\overline{x}[/itex], [itex]\overline{y}[/itex], and/or [itex]\overline{xy}[/itex] are wrong or [itex]\overline{x}\overline{y}[/itex] is NOT equal to [itex]\overline{xy}[/itex].

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

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

Using that definition of [itex]\overline{u}[/itex] 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

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

Show that [itex]\overline{xy}=\overline{x}.\overline{y}[/itex]

Homework Equations

[itex]\overline{x}=\int^{∞}_{-∞}{x.f(x)dx}[/itex]

The Attempt at a Solution

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

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

But when it comes to [itex]\overline{xy}[/itex] 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):
[tex] 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. [/tex]
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}##.