1
$\begingroup$

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 $a_i$?

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – whuber
    Commented Jan 21, 2022 at 15:25

1 Answer 1

0
$\begingroup$

Solution for probability distribution

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 $c$

$p(x)= \frac{1}{2\pi c} \int_{\mathbf{R}} e^{i\frac{s}{c}x} \overline{\varphi_{Z}(s)}{\rm d}s$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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