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

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    $\begingroup$ Apply whatever definition of expectation you're using. Eg. $\mathrm{E}[Y] = \sum_\omega Y(\omega) P(\omega)$ or $\mathrm{E}[Y] = \int_\omega Y(\omega) dP(\omega)$. $\endgroup$ – Matthew Gunn Oct 14 '16 at 1:58
  • $\begingroup$ Expectation is an integral. Write the integral and use its linearity properties. $\endgroup$ – Glen_b Oct 14 '16 at 3:04

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.

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    $\begingroup$ why is the pmf p(x) within the sum, and not p(ax+b) ? $\endgroup$ – sousben Oct 4 '18 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 '19 at 20:17

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