Finding covariance using the joint pdf

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The discussion revolves around calculating the covariance Cov(X,Y) using the joint probability density function f(x,y) = 24xy for the region defined by 0 < x + y < 1. The user initially applied Cov(X,Y) = E(XY) - E(X)E(Y) but did not arrive at the expected answer of 1/15. A suggestion was made to use an alternative definition of covariance, which involves calculating (X - E[X])(Y - E[Y])f(x,y)dxdy to verify the calculations. After re-evaluating using this method, the user found Cov(X,Y) = -2/75, indicating a consistent result but not matching the suggested answer. The conversation emphasizes the importance of careful step-by-step calculation in probability and statistics.
Flashcop
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The following question appeared on a practice exam:

For
f(x,y) = 24xy if 0<x+y<1 , 0<x,y
0 elsewhere

find Cov(X,Y)

I used Cov(X,Y) = E(XY) - E(X)E(Y) to calculate covariance, with

E(XY) = \int^{1}_{0}\int^{1-y}_{0}24x^{2}y^{2}dxdy

but for some reason I didn't get the suggested answer of 1/15

Can someone explain to me what I did wrong? Was there something wrong with the terminals?
 
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Flashcop said:
The following question appeared on a practice exam:

For
f(x,y) = 24xy if 0<x+y<1 , 0<x,y
0 elsewhere

find Cov(X,Y)

I used Cov(X,Y) = E(XY) - E(X)E(Y) to calculate covariance, with

E(XY) = \int^{1}_{0}\int^{1-y}_{0}24x^{2}y^{2}dxdy

but for some reason I didn't get the suggested answer of 1/15

Can someone explain to me what I did wrong? Was there something wrong with the terminals?

Hey Flashcop and welcome to the forums.

If I were you I would in later cases use the alternative definition COV(X,Y) = E[[X-E[X][Y-E[Y]] and calculate (X - E[X])(Y - E[Y])f(x,y)dxdy for the integral to double check your calculations if you think something is awry (it is redundant but it is a good way to cross-check your calculations in the case of say an algebraic mistake).

For this problem, I think the best way to get relevant advice is to go through your calculation step by step and post it here so that we can go through your algebra and your reasoning.
 
chiro said:
Hey Flashcop and welcome to the forums.

If I were you I would in later cases use the alternative definition COV(X,Y) = E[[X-E[X][Y-E[Y]] and calculate (X - E[X])(Y - E[Y])f(x,y)dxdy for the integral to double check your calculations if you think something is awry (it is redundant but it is a good way to cross-check your calculations in the case of say an algebraic mistake).

For this problem, I think the best way to get relevant advice is to go through your calculation step by step and post it here so that we can go through your algebra and your reasoning.

Thanks mate :)

So using the alternate definition, I got:

Cov(X,Y) = 24 \int^{1}_{0}\int^{1-y}_{0}(x-\frac{2}{5})(y-\frac{2}{5})xydxdy
= 24 \int^{1}_{0}\int^{1-y}_{0}x^{2}y^{2}-\frac{2x^{2}y}{5}-\frac{2xy^{2}}{5}+\frac{4xy}{25}dxdy
=24\int^{1}_{0}\frac{-1}{75}(2-5y)^{2}(y-1)^{2}ydy
=\frac{-24}{75}\int^{1}_{0}(2-5y)^{2}(y-1)^{2}ydy
=\frac{-24}{75}*\frac{1}{12}<br /> =\frac{-2}{75}

Which is the same answer I got before, so that probably rules out any algebraic errors.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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