This seems rather simple. I know that the expected value of a constant is just the constant. But I feel like I'm missing something. Any help would be appreciated, thanks

  • 6
    $\begingroup$ Apply whatever definition of expectation you're using. Eg. $\mathrm{E}[Y] = \sum_\omega Y(\omega) P(\omega)$ or $\mathrm{E}[Y] = \int_\omega Y(\omega) dP(\omega)$. $\endgroup$ Oct 14, 2016 at 1:58
  • $\begingroup$ Expectation is an integral. Write the integral and use its linearity properties. $\endgroup$
    – Glen_b
    Oct 14, 2016 at 3:04

1 Answer 1


Suppose X is a discrete random variable with pmf $p(x)$. Then, by definition, \begin{eqnarray*} E(aX+b)&=& \sum_{x}(ax+b)p(x)\\ &=&\sum_{x}(ax\cdot p(x)+b\cdot p(x))\\ &=&\sum_{x}ax\cdot p(x) + \sum_{x}b\cdot p(x)\\ &=&a\underbrace{\left(\sum_{x}x\cdot p(x)\right)}_{E(X)} + b\underbrace{\left(\sum_{x}p(x)\right)}_{1}\\ &=&a\cdot E(X) + b \end{eqnarray*} Similarly, the result can be obtained when $X$ is a continuous random varaible.

  • 1
    $\begingroup$ why is the pmf p(x) within the sum, and not p(ax+b) ? $\endgroup$
    – sousben
    Oct 4, 2018 at 21:57
  • $\begingroup$ Good question. Because expected value $E(X)$ does not behave like a regular function as far as substitution goes. The definition of expected value is $E(X) = \sum_{x}x\cdot p(x) $ or more generally $E(f(X)) = \sum_{x}f(x)\cdot p(x)$, where $x$ ranges over the domain of $X$ in both definitions. $\endgroup$
    – john
    Jul 15, 2019 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.