 # If I know the characteristic function phi_X(t) of a random variable X>0, how can I write the characteristic function phi_Y(t) of Y=log(X)? Elisabeth Wiley 2022-08-12 Answered
Characteristic function of logarithm of random variable
If I know the characteristic function ${\varphi }_{X}\left(t\right)$ of a random variable $X>0$, how can I write the characteristic function ${\varphi }_{Y}\left(t\right)$ of $Y=\mathrm{log}\left(X\right)$?
I know that ${\varphi }_{X}\left(t\right)=E\left[{e}^{itX}\right]$ and ${\varphi }_{Y}\left(t\right)=E\left[{e}^{it\mathrm{log}\left(X\right)}\right]$. But I can't derive one from the other. Any idea? I would like to use ${\varphi }_{X}\left(t\right)$ to calculate the second moments $X$
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This question comes up in various guises: knowing the Fourier transform of $f$ (in your case, the probability density function of $X$), can we find the Fourier transform of another related function (in your case, the p.d.f. of $\mathrm{log}X$)? Unfortunately, nonlinear transformations completely mess up the picture of the Fourier transform. All one can do is to invert the Fourier transform, apply the desired nonlinear transformation, and take the Fourier transform of that.

We have step-by-step solutions for your answer! Gauge Roach
We need to compute the log-characteristic function that is the characteristic of $Y=\mathrm{ln}\left(X\right)$. So, we need to express the characteristic function of $Y$ in terms of values $X$:
$E\left(\mathrm{exp}\left(ikY\right)\right)={\int }_{{l}_{y}}^{{u}_{y}}\mathrm{exp}\left(iky\right){f}_{Y}\left(y\right)dy$
As $y=\mathrm{ln}\left(x\right)$, we have $dy=\frac{dx}{x}$ and so :
$E\left(\mathrm{exp}\left(ikY\right)\right)={\int }_{\mathrm{exp}\left({l}_{y}\right)}^{\mathrm{exp}\left({u}_{y}\right)}\mathrm{exp}\left(ik\mathrm{ln}\left(x\right)\right){f}_{Y}\left(\mathrm{ln}\left(x\right)\right)\frac{dx}{x}$
We need to express the density ${f}_{Y}$ of $Y$ in terms of the density ${f}_{X}$ of $X$. As $Y=g\left(X\right)=\mathrm{ln}\left(X\right)$, because the probabilities are invariant by change of variables, the transformation is strictly monotone and derivable in all the domain of X so that:
${f}_{Y}\left(y\right)=|\frac{1}{{g}^{\prime }\left({g}^{-1}\left(y\right)\right)}|.{f}_{X}\left({g}^{-1}\left(y\right)\right)$
$⇔{f}_{Y}\left(y\right)=\mathrm{exp}\left(y\right).{f}_{X}\left(\mathrm{exp}\left(y\right)\right)=x.{f}_{X}\left(x\right)$
In the end, the log characteristic function equals:
$E\left(\mathrm{exp}\left(ikY\right)\right)={\int }_{\mathrm{exp}\left({l}_{y}\right)}^{\mathrm{exp}\left({u}_{y}\right)}\mathrm{exp}\left(ik\mathrm{ln}\left(x\right)\right)x.{f}_{X}\left(x\right)\frac{dx}{x}$
$⇔E\left(\mathrm{exp}\left(ikY\right)\right)={\int }_{\mathrm{exp}\left({l}_{y}\right)}^{\mathrm{exp}\left({u}_{y}\right)}\mathrm{exp}\left(ik\mathrm{ln}\left(x\right)\right){f}_{X}\left(x\right)dx$
Q.E.D.

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