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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 $a_i$?

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  • $\begingroup$ 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
    Jan 21, 2022 at 15:25

1 Answer 1

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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 $c$

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

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