1
$\begingroup$

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 $W$?

$\endgroup$
2
  • $\begingroup$ 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 $X$. $\endgroup$ Commented Nov 21, 2015 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. $\endgroup$ Commented Nov 21, 2015 at 17:59

1 Answer 1

Reset to default
1
$\begingroup$

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 $V(W) = \frac{1}{4}V(Y)$.

$\endgroup$
2
  • $\begingroup$ Thank you for the explanation. I'll use Variance formula you gave to read how this works in depth. Many thanks. $\endgroup$ Commented Nov 21, 2015 at 17:57
  • $\begingroup$ This explains how the Variance formula is obtained. $\endgroup$ Commented Nov 21, 2015 at 18:09

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.