Let us assume that $X$ is a random variable and $a$ is a constant. Now suppose $Y=a+bX$, what would the characteristic function of $Y$ would be? Is it? \begin{eqnarray} \mathbb{E}_X\left[\exp(iuY)\right]&=&\mathbb{E}_X\left[\exp(iu\{a+bX\})\right]\\ &=&\mathbb{E}_X\left[\exp(iua)\exp(iubX)\right] \end{eqnarray} How can this be decomposed further? What is the intuition behind the characteristic function of a constant $a$? Would it be wrong to further decompose the above as \begin{equation} \mathbb{E}_X\left[\exp(iuY)\right]\overset{?}{=}\exp(iua)\mathbb{E}_X\left[\exp(iubX)\right] \end{equation} given that $a$ is a constant? Not even sure that is correct. Thanks!

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    $\begingroup$ Are you aware that expectation is a linear functional? If not, that's the very first thing to study next. $\endgroup$ – whuber Dec 18 '19 at 0:05
  • $\begingroup$ @whuber I am not sure I follow. Yes if we have $\mathbb{E}(X+Y)$ that can be expressed as $\mathbb{E}(X)+\mathbb{E}{(Y)}$, but if we have $\mathbb{E}(\exp(X+Y))$ that is no longer the case, as it can only be decomposed to $\mathbb{E}(\exp(X)\exp(Y))$. Could you please elaborate further? Perhaps I am missing something very obvious. $\endgroup$ – Carl Dec 18 '19 at 0:13
  • $\begingroup$ Linear means that for any numbers $a_1,a_2,\ldots,a_n$ and any random variables $X_1,X_2,\ldots, X_n,$ $$\mathbb{E}(a_1X_1+a_2X_2+\cdots+a_nX_n)=a_1\mathbb{E}(X_1)+a_2\mathbb{E}(X_2)+\cdots+a_n\mathbb{E}(X_n).$$ Apply this to the case $n=1$ and $a_1=\exp(iua)$ (which is a number). $\endgroup$ – whuber Dec 18 '19 at 14:27
  • $\begingroup$ @whuber Thank you for elaboration. I am aware of the above and hence my final derivation. In that case that implies that my last decomposition correct. Right? I was conflicted with this result, as I wasn't sure about the intuition of $\exp(iua)$ without an expectation operator. $\endgroup$ – Carl Dec 18 '19 at 14:31
  • $\begingroup$ It's just a number. $\endgroup$ – whuber Dec 18 '19 at 14:32

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