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Let the joint probabilty mass function of discrete random variables X and Y be given by

$f(x,y)=\frac{x^2+y^2}{25}$, for $(x,y) = (1,1), (1,3), (2,3)$

The value of E(Y) is ?

Attempt

$E(Y) = \sum_{x,y} y\cdot\frac{x^2 + y^2}{25}$

$E(Y) = \sum_{x,y}\frac{x^2y + y^3}{25}$

Substituting for $(x,y) = (1,1), (1,3), (2,3)$

$E(Y) = \frac1{25} + \frac{30}{25} + \frac{39}{25}$

$E(Y) = 2.80$

Is this right?

RStyle
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  • There's a 2/25 chance y=1, and it's 3 in all other cases, so (2/25)1+(23/25)3 = 2.84 (or 71/25) is what I got. –  Oct 08 '16 at 16:18

2 Answers2

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\begin{align} \mathbb E[Y] &= \sum_y y\cdot\mathbb P(Y=y)\\ &= 1\cdot\mathbb P(Y=1) + 3\cdot\mathbb P(Y=3)\\ &= \frac{1^2+1^2}{25} + 3\left(\frac{1^2+3^2}{25}+\frac{2^2+3^2}{25} \right)\\ &= \frac{71}{25}. \end{align}

Math1000
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In relation to the (1,1) point you seem to be claiming that $1\cdot \frac{1^2+1^2}{25}=\frac{1}{25}$.

It's more usually thought to be the case that $1^2+1^2>1$*. This seems to be the cause of your problem here.

* (Some people - reckless people perhaps - even claim that $1^2+1^2$ could be as much as $2$.)

Glen_b
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