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Let $X$ be a random variable, and let $\phi_X(t)$ be its characteristic function. Let $Y = f(X)$ be a transformation of the random variable $X$ where $f$ is increasing and one-to-one. Is there a direct functional relationship between the characteristic function of the transformation $f(X)$ and the characteristic function $\phi_X(x)$? Meaning if $\phi_Y(t)$ is the characteristic function of the random variable $f(X)$, can we write $\phi_Y(t)$ in terms of $\phi_X(t)$?

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  • $\begingroup$ I think if there was any relation, it'd be quite well known. fourier transforms don't have special behaviour under transformations. $\endgroup$ – Lost1 Jan 20 '14 at 20:21
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If $Y=g(X)$ (let me use $g$, reserving $f$ for the densities) :

Then $$ \phi_Y(t)=E(e^{i Yt})=\int_{-\infty}^{\infty} f_Y(y) e^{i y t} dy= \int_{-\infty}^{\infty} f_X(x) e^{i g(x) t} dx=\\ =\frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \phi_X(s) e^{-i s x } ds \, e^{i g(x) t} dx $$

Of course, this is rather trivial ( $ \phi_X(t) \to f_X(x) \to f_Y(y) \to \phi_Y(t)$) and formal (the formula is not practical, it does not give a nice or simple relationship between $\phi_Y(t)$ and $\phi_X(t)$ but I don't see why you'd expect that it can be simplified.

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Here's I think what would be a counterexample: if $X \sim R[0,1]$ and $Y \sim N(X, \sigma^2)$, then MGF of $X$ would be $\frac{e^t-1}{t}$ and MGF of $Y$ would be $e^{xt+\frac{\sigma^2 t^2}{2}}$ and I doubt there's a way to express the latter via the former.

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  • $\begingroup$ What about something simpler. Consider $X \sim N(\mu , \sigma)$ and consider $f(x)= x/2$. (Even you could make it simpler by taking $\mu=0$ and $\sigma=1$). $\endgroup$ – kolonel Jan 20 '14 at 3:31
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    $\begingroup$ This was meant as a counterexample. I didn't say it's impossible for some functions. $\endgroup$ – Alex Jan 20 '14 at 3:35
  • $\begingroup$ Thanks. But in your case $Y=f(X)$ is not increasing and one-to-one or am I missing something? $\endgroup$ – kolonel Jan 20 '14 at 3:49
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This is more of a comment than answer and is for someone stumbling into this post.

For linear transformations of the form $Y=aX+b$, we can say that $\phi_Y(t)=e^{itb}\phi_X(at)$. If we are dealing with random vectors, then for transformation $AX+B$, we have $\phi_Y(t)=e^{it^\top B}\phi_X(A^\top t)$.

Source: https://www.statlect.com/fundamentals-of-probability/joint-characteristic-function and references therein.

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  • $\begingroup$ please use the comment function for comments. $\endgroup$ – Fabian Rost May 9 '17 at 7:33

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