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When provided a density function and after solving for the expected value E(Y) and the variance V(Y), I am trying to understand why finding the expected value of a dollar value function seemingly works differently than finding the variance of the dollar value function.

For example, the density function is $$ f(y) = \begin{cases}\frac{3}{2}y^2 + y \quad \quad 0 \leq y \leq 1 \\ 0 \quad \quad \text{elsewhere} \end{cases}$$ This gives $ E(Y) = .7083$ and $V(Y) = .0483$.

If the dollar value is given by $W = 5 - .5Y$, to find the mean of $W$, simply solving for $W = 5 - .5[E(Y)]$ gives the correct answer.

However, it does not seem to work the same for finding the variance of $W$, as it is incorrect to say that $V(W) = 5 - .5[V(Y)]$. What is the difference between finding the mean and variance in this case, and how can I understand how to solve for the variance of $W$?

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  • $\begingroup$ Part answer: Let $Y=X+k$. Then the variance of $Y$ is $E((Y-\mu_Y)^2)$. But $Y=X+k$ and $\mu_Y=\mu_X+k$, so $Y-\mu_Y=X-\mu_X$, which means the $k$ has disappeared. The "wiggliness" of $X+k$ is the same as the wiggliness of $X$. $\endgroup$ Nov 21 '15 at 17:50
  • $\begingroup$ @AndréNicolas, thank you for the response. Your answer is a bit over my head, but that has more to do with me than your answer. I'll work on understanding this. $\endgroup$ Nov 21 '15 at 17:59
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The expectation is linear - that is, $E(aX + bY) = aE(X) + bE(Y)$ for all $a,b\in \mathbb{R}$ and $X,Y$. In particular

$$E(W) = E(5) - \frac{1}{2}E(Y) = 5 - \frac{1}{2}E(Y)$$

However, the variance is not linear (unless the random variables are uncorrelated). Instead, it satisfies

$$ V(aX + b) = a^2V(X) $$

for all $a,b\in \mathbb{R}$ and $X$. Thus, we have $V(W) = \frac{1}{4}V(Y)$.

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  • $\begingroup$ Thank you for the explanation. I'll use Variance formula you gave to read how this works in depth. Many thanks. $\endgroup$ Nov 21 '15 at 17:57
  • $\begingroup$ This explains how the Variance formula is obtained. $\endgroup$ Nov 21 '15 at 18:09

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