# How can we write the below characteristic function?

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 $$$$\mathbb{E}_X\left[\exp(iuY)\right]\overset{?}{=}\exp(iua)\mathbb{E}_X\left[\exp(iubX)\right]$$$$ given that $$a$$ is a constant? Not even sure that is correct. Thanks!

• Are you aware that expectation is a linear functional? If not, that's the very first thing to study next. – whuber Dec 18 '19 at 0:05
• @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. – Carl Dec 18 '19 at 0:13
• 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). – whuber Dec 18 '19 at 14:27
• @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. – Carl Dec 18 '19 at 14:31
• It's just a number. – whuber Dec 18 '19 at 14:32