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In my exercise, I'm given two independent discrete stochastic variables, with the probability function;

$$ p_{X}\left(x\right)=\left\{ \begin{array}{cc} \frac{1}{4} & \text{if } x = -1 \\ \frac{1}{4} & \text{if } x = 0 \\ \frac{1}{2} & \text{if } x = 1 \\ 0 & \text{otherwise} \end{array} \right. $$

I'm given the task to find several properties, for instance the mean ($EX = \frac{1}{4}$), and variance ($VarX=\frac{11}{16}$). I was able to do this with any issues.

Now I'm given the task of calculating, the variance; $Var(X-5Y)$, based upon the fact that independent variables have a covariance of 0, I've found this to be; $Var(X-5Y)=\frac{143}{8}$.

However when I'm trying to confirm that the covariance is 0, I get in trouble; $$ Cov(X,Y) = E(XY) - EX \cdot EY $$ With; $$E(XY) = \left(\frac{1}{4} \cdot -1 \cdot -1 \right) + \left(\frac{1}{4} \cdot 0 \cdot 0 \right) + \left(\frac{1}{2} \cdot 1 \cdot 1 \right) = \frac{3}{4}$$ I'm getting; $$ Cov(X,Y) = \frac{3}{4} - \frac{1}{4} \cdot \frac{1}{4} = \frac{11}{16} $$ Clearly I'm doing something wrong, but what?

Skeen
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  • $X$ and $Y$ are independent, so $E(XY) = \left(\frac{1}{2^2} + \frac{1}{4^2}\right)\cdot 1 + \left(\frac{1}{2}\frac14 + \frac14\frac12\right)\cdot (-1) = \frac{5}{16} - \frac{4}{16}$. – Daniel Fischer Aug 08 '13 at 15:19
  • What would the general formula for E(XY) be? $E\left(X_{1},X_{2}\right) = \sum_{i \in I} x_{i1} x_{i2} p\left(x_{i1},x_{i2}\right)$? – Skeen Aug 08 '13 at 15:35
  • $\sum P(X=x)P(Y=y)xy$ for the finite or countable case. $P(X=x)P(Y=y) = P\bigl((X=x) \land(Y=y)\bigr)$ by independence. – Daniel Fischer Aug 08 '13 at 15:37
  • What would $E(XY)$ look, extended, with the finite case formula? - And how would that differ, from what I've got, clearly it should yield what you wrote, as first comment. Is my support ${(-1,-1),(-1,1), (1,-1), (1,1)}$? – Skeen Aug 08 '13 at 15:42
  • Okay, so it seems like the support has to be ${(-1,-1),(-1,1), (1,-1), (1,1)}$, is it always the case, that the support for this, is the Cartesian product of the normal support? – Skeen Aug 08 '13 at 15:48
  • $$\sum_{(x,y) \in \Omega_X\times\Omega_Y} P(X=x,Y=y)\cdot x\cdot y$$ That's the general definition of $E(XY)$. Here, both have the same possible outcomes, and are independent, so we obtain $$E(XY) = \sum_{y\in \Omega}\left(\sum_{x\in\Omega} P(X=x,Y=y)x\right)y = \sum_{y\in\Omega}P(Y=y)\left(\sum_{x\in\Omega}P(X=x)x\right)y = E(X)E(Y).$$ – Daniel Fischer Aug 08 '13 at 15:49
  • @DanielFischer: If you'd like, turn your comments into an answer, and I'll accept that. – Skeen Aug 08 '13 at 23:16

1 Answers1

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The expected value of the product of two random variables is obtained by summing over all possible pairs of outcomes,

$$E(XY) = \sum_{y \in \Omega_Y}\sum_{x\in\Omega_X} P(X=x,Y=y)\cdot x\cdot y.$$

In the particular case of two independent random variables, the joint probability of a pair is the product of the probabilities of the components,

$$P(X=x,Y=y) = P(X=x) \cdot P(Y=y),$$

and then the sum splits

$$\begin{align}E(XY) &= \sum_{y \in \Omega_Y}\sum_{x\in\Omega_X} P(X=x)P(Y=y)\cdot x\cdot y\\ &= \sum_{x\in\Omega_X} P(X=x)\cdot x \cdot \sum_{y\in\Omega_Y} P(Y=y)\cdot y\\ &= E(X)\cdot E(Y). \end{align}$$

For two inependent identically distributed variables with the given distribution, that is

$$\begin{align}E(XY) &= P(X=1,Y=1)\cdot1 + P(X=-1,Y=-1)\cdot 1\\ &\qquad + P(X=1,Y=-1)\cdot(-1) + P(X=-1,Y=1)\cdot (-1)\\ &= \left(\frac{1}{2^2} + \frac{1}{4^2}\right)\cdot 1 + \left(\frac12\frac14 + \frac14\frac12\right)\cdot (-1)\\ &= \frac{1}{16} \end{align}$$

by the explicit summation (ignoring $0$ terms).

Daniel Fischer
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