# Characteristic function of linearly transformed random variables with extracted factor

Given is a sequence of independent random variables $$X_1, X_2,\ldots, X_n$$ with characteristic function $$\varphi_{X_i}(t)$$. The characteristic function of $$Y = \sum_{i=1}^n a_i X_i$$, where the $$a_i$$ are constants, is given by $$\varphi_{Y}(t)=\varphi_{X_1}(a_1t)\varphi_{X_2}(a_2t)\cdots \varphi_{X_n}(a_nt)$$. The probability distribution of $$Y$$ is $$p(x)= \frac{1}{2\pi} \int_{\mathbf{R}} e^{itx} \overline{\varphi_{Y}(t)}{\rm d}t$$.

Question

Let $$a_i=c b_i$$ and $$Z = c\sum_{i=1}^n b_i X_i$$. Is it possible to express the probability distribution and characteristic function of $$Z$$ if we extract a common factor $$c$$ from the constants $$a_i$$?

• It looks like you can skip all the stuff about characteristic functions. Because $Z=Y,$ there's nothing to ask or answer. You probably meant to define $Z=\sum b_i X_i = Y/c.$ Given any expression for the distribution of $Y,$ though, writing a comparable expression for the distribution of $Y/c$ is completely elementary. Are you sure you have stated the question you actually have?
– whuber
Jan 21, 2022 at 15:25

In the original characteristic function $$\varphi_{Z}(t)=\varphi_{X_1}(cb_1t)\varphi_{X_2}(cb_2t)\cdots \varphi_{X_n}(cb_nt)$$ we set $$s=ct$$ and get $$\varphi_{Z}(s)=\varphi_{X_1}(b_1s)\varphi_{X_2}(b_2s)\cdots \varphi_{X_n}(b_ns)$$. Then we replace in the original probability distribution $$t=\frac{s}{c}$$ and $${\rm d}t=\frac{{\rm d}s}{c}$$ and can express the probability distribution using the extracted factor $$c$$
$$p(x)= \frac{1}{2\pi c} \int_{\mathbf{R}} e^{i\frac{s}{c}x} \overline{\varphi_{Z}(s)}{\rm d}s$$.