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So I have the following tables

$$ \left[ \begin{array}{c|ccc} x&-3&6&9\\ f(x)&\frac{1}{6}&\frac{1}{2}&\frac{1}{3} \end{array} \right] $$

I am tasked to find the following values

E(X), E(X2),E(4X2 + 4X + 1)

For E(X) I got 5.5
For E(X2) I got 46.5
And for E(4x2+4x+1) I got 61

Are these values correct? They seem sort of high to me,

Thanks John

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  • $\begingroup$ Looks much too low. $\endgroup$ – André Nicolas Jul 12 '16 at 4:23
  • $\begingroup$ $(2X+1)^2$ takes values $25, 169, 361$ with probabilities $1/6,1/2,1/3$. Now calculate. $\endgroup$ – André Nicolas Jul 12 '16 at 4:27
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By Linearity of expectation, we have $$\mathbb{E}[4X^2+4X+1] = 4\mathbb{E}[X^2]+4\mathbb{E}[X]+1.$$

If you plug in $\mathbb{E}[X] = 5.5$, $\mathbb{E}[X^2] = 46.5$, and $\mathbb{E}[4X^2+4X+1] = 61$, the left side of the above equation is $61$, but the right side is $4 \cdot 46.5 + 4 \cdot 5.5 + 1 = 209$. So you have a mistake somewhere.

I'll go ahead and tell you that your answers for $\mathbb{E}[X]$ and $\mathbb{E}[X^2]$ are correct: $$\mathbb{E}[X] = \dfrac{1}{6} \cdot -3 + \dfrac{1}{2} \cdot 6 + \dfrac{1}{3} \cdot 9 = 5.5$$ $$\mathbb{E}[X^2] = \dfrac{1}{6} \cdot (-3)^2 + \dfrac{1}{2} \cdot 6^2 + \dfrac{1}{3} \cdot 9^2 = 46.5$$

So you should double check your calculations on the last part.

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  • $\begingroup$ Duh, math is hard when you're tired lol $\endgroup$ – John Hamlett IV Jul 12 '16 at 4:32
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The first and second look good, but the third can't be right.

Since expectation is linear in its argument ($E(aX + b) = aE(X) + b$),

$$ E(4X^2 + 4X + 1) = 4E(X^2) + 4E(X) + 1 = 4(46.5) + 4(5.5) + 1 = 209. $$

(You should double check this by calculating the third expectation directly again.)

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