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Let the joint random variable $P[x;y]$ be

$P[x;y] = c[2x^2 + y^2], x=-1;0;1, y=1;2;3;4$

$=0$ $elsewhere$

So I had to find the value of $c$ that makes $P[x;y]$ a joint discrete random variable.

I think I did that right. I just add up all the probabilities where $x = -1;0;1$ and $y = 1;2;3;4$ and made it equal to 1. Then I solved for $c$ and got $c=\frac{1}{106}$. Please check this for me if you think I've done something wrong.

Now they asked to calculate the $E[y]$ and the next question asked if $x$ and $y$ are independent.

So. I was a bit confused with the whole $P[x;y]$ and how to split it up into $P[x]$ and $P[y]$ (if that's possible) and then how to work out the $E[y]$.

StephanCasey
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  • So you got the normalisation correct (barring mental arithmetic error) but your approach was correct..now you have to find the marginal probabilitys this done by summing over all x to get the y distribution, and similarly for the x distribution you sum over y. Then you take expectations with respect to the marginals. To get independence, I suggest going through notes that you have in terms of what criterion is required for that :). – Chinny84 Sep 08 '14 at 16:04
  • I'm seeing something that looks like $E[Y] = \sum _{x} [{y}P[x;y]]$ – StephanCasey Sep 08 '14 at 16:09
  • That would work. Presumably this is an excerise to understand the joint distributions, I would advice derive the marginals first $$P(y) = \sum_{x}P(x,y)$$ and then compute the expectation for clarity..though your approach would work (skipping a few steps :)) – Chinny84 Sep 08 '14 at 16:14
  • I don't really understand the notation though. If we're only summing x's then which y values do we use? – StephanCasey Sep 08 '14 at 16:16
  • sorry I was not clear before. you obtain the marginal probability for y as I shown above. Then you sum over all y after multiplying the marginal probability of y, $P(y)$, by y. Does that make sense? – Chinny84 Sep 08 '14 at 16:18
  • I think what I'm confused about is how to I use that formula to calculate the marginal probability of y. Could you expand it for me? What I mean is. $P(y) = \sum _x P(x,y) = (-1, y) + (0,y) + (1,y)$ because it said sum of x's but I feel like I don't understand the notation – StephanCasey Sep 08 '14 at 16:19

3 Answers3

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$$E(Y)=\sum_{x,y}y\,P_{X,Y}(x,y)=\sum_yy\,P_Y(y),\qquad P_Y(y)=\sum_xP_{X,Y}(x,y)$$

Did
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To clarify the points above $$ \begin{align} P_Y(y) &=& \sum_{x = \lbrace-1,0,1\rbrace} c\left(2x^2 + y^2\right)\\ &=&c\left[\sum_{x = \lbrace-1,0,1\rbrace} 2x^2 + \sum_{x = \lbrace-1,0,1\rbrace} y^2\right]\\ &=&c\left[2\cdot (-1)^2+ 2\cdot(0)^2+2\cdot(1)^2 + 3\cdot y^2\right] \\ &=& c\left(4+3y^2\right) \end{align} $$

now you can apply the formula as @Did.

Chinny84
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  • Oh thanks. Now it's making more sense to me – StephanCasey Sep 08 '14 at 16:29
  • so just to clarify. I can say. $E(y) = \sum_{y}{y}{\frac{1}{106}}({4 + 3y^2}) = (1)(\frac{1}{106})(4+3(1)^2) + (2)(\frac{1}{106})(4+3(2)^2) + (3)(\frac{1}{106})(4+3(3)^2) + (4)(\frac{1}{106})(4+3(4)^2)$ – StephanCasey Sep 08 '14 at 16:40
  • that seems ok to my eye there. – Chinny84 Sep 08 '14 at 16:42
  • Thanks. So I suppose the place where I was confused is just that you only need to use the y's in the equation since you eliminated the x's when working out $P_y(y)$ – StephanCasey Sep 08 '14 at 16:45
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    yes, and as Did has shown the link between computing the expectation of Y with the joint distribution, and how that relates to the marginal :). – Chinny84 Sep 08 '14 at 16:47
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I have worked out the problem for you and I hope it helps to check with your answer. I have done it in EXCEL what other responders have alluded to. To verify your answer.enter image description here

Satish Ramanathan
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