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This is a very trivial question but I have a lot of trouble dealing with Expectation, summation and product together. So I need find this:

$\mathbb{E}(\sum_{i=1}^n m_i X_i)$

where $m_i$ are constants.

I know that next step would be:

$\sum_{i=1}^n \mathbb{E}(m_i X_i)$

After this, would the next step be:

$\sum_{i=1}^n m_i . \mathbb{E}(X_i)$

or

$\sum_{i=1}^n m_i . \sum_{i=1}^n\mathbb{E}(X_i)$

?

Thank you.

weak_at_math
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    Expectation is a linear operator. – Hamdiken Mar 03 '23 at 23:42
  • Everything you did except the last step looks fine. Expectation is a linear operator so $\mathsf E(\alpha X+\beta Y)=\alpha\mathsf E(X)+\beta\mathsf E(Y)$. This generalizes to a sum of many terms. – Aaron Hendrickson Mar 03 '23 at 23:42
  • $\sum_{i=1}^n m_i . \mathbb{E}(X_i)\neq \sum_{i=1}^n m_i . \sum_{i=1}^n\mathbb{E}(X_i)$ – Aaron Hendrickson Mar 03 '23 at 23:44
  • You can try this out yourself if it's confusing (as it can be). Suppose $n = 2$ and you have $X_1$ a random variable that is always equal to $2$ (this is still a random variable—it's just a constant one!) and $X_2$ a random variable that is always equal to $3$. Let $m_1 = 5, m_2 = 7$. (These numbers are just picked out of a hat; there's nothing special about them.) Which of your two choices is equal to your second-to-last step? – Brian Tung Mar 04 '23 at 00:17
  • Another way to think about it (which is probably more useful after you've done the previous little "exercise"): Does $m_1$ "know about" $X_2$? Does $m_2$ know about $X_1$? – Brian Tung Mar 04 '23 at 00:19

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Possibly the best way to think about this is to start with a simple example. If $n = 2$, then $\sum_{i = 1}^2 m_i X_i = m_1 X_1 + m_2 X_2$. Then if we look at the expectation of it, we get

$\begin{eqnarray}\mathbb{E}(\sum_{i = 1}^2 m_i X_i) & = & \mathbb{E}(m_1 X_1 + m_2 X_2) \\ & = & \mathbb{E}(m_1 X_1) + \mathbb{E}(m_2 X_2) \\ & = & m_1 \mathbb{E}(X_1) + m_2 \mathbb{E}(X_2) \end{eqnarray}$

you should be able to simplify that last line into a sigma form, and from there you can intuit what happens with more general $n$.

ConMan
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