Plot the density function of a normal random variable knowing only the characteristic function in R

Question

The characteristic function of a normal random variable with mean $\mu$ and standard deviation $\sigma$ is:

$$\begin{alignat*}{1} \hat{\phi}(t) & =e^{i\mu t}e^{-\frac{1}{2}\sigma^{2}t^{2}}\\ & =\exp(i\mu t-\frac{1}{2}\sigma^{2}t^{2}) \end{alignat*}$$

What I want to do is to plot the PDF of the normal random variable knowing only the characteristic function outlined above $\hat\phi(t)$. Ideally the solution to this problem would be to apply the Inverse Fourier Transform on $\hat\phi(t)$ and then plot the result for different values of $x$. Applying the Inverse Transform leads me to solve this integral to get the PDF of $x$:

$$\begin{alignat*}{1} f(x) & =\frac{1}{2 \pi}\int_{-\infty}^{\infty}\hat{\phi}(t)\exp(-itx)dt\\ & =\frac{1}{2 \pi}\int_{-\infty}^{\infty}\exp(i\mu t-\frac{1}{2}\sigma^{2}t^{2})\exp(- itx)dt\\ & =\frac{1}{2 \pi}\int_{-\infty}^{\infty}\exp(i\mu t-\frac{1}{2}\sigma^{2}t^{2}- itx)dt \end{alignat*} $$

The problem is that I do not know how to do it in R. I could use the integrate() function, but how do I treat the imaginary unit $i$?

Answer 1

Just read off the coefficients of $t$ and $t^2$ in the characteristic function in order to deduce the values of $\mu$ and $\sigma^2$. If the characteristic function is not (or cannot be put) in the form $exp(a i t - b t^2)$, where $a$ is a real number and $b$ is a non-negative real number, then it is not the characteristic function of a Normal random variable. If it is of that form, then $\mu = a$ and $\sigma^2 = 2 b$. Now plot a Normal density with those parameters to your heart's content.

Your formula for the probability density function is incorrect. it should be $$\begin{alignat*}{1} f(x) & =\frac{1}{2 \pi}\int_{-\infty}^{\infty}\hat{\phi}(t)\exp(-itx)dt\\\end{alignat*}$$which lo and behold, comes out correct.