Characteristic function of $Z = X + \sigma Y$

Question

Let $X$ and $Y$ be independent, assume that $Y$ has $\mathcal{N}(0,1)$ distribution. Let $\sigma >0$ and let $\phi$ be the characteristic function of $X: \phi(u) = \mathbb{E}[e^{iuX}]$.

Now, we study the density function of the random variabel $Z=X+\sigma Y$. The aim is to show that the density $p(z)$ could be expressed as \begin{align} &p(z) = \frac{1}{2\pi\sigma} \int_\mathbb{R} \phi\big(\frac{-y}{\sigma}\big)\exp\big(\frac{iyz}{\sigma - \frac{1}{2}y^2}\big)\ dy.\\ \textbf{Edit: this should be} \qquad &p(z)= \frac{1}{2\pi\sigma} \int_\mathbb{R} \phi\big(\frac{-y}{\sigma}\big)\exp\big(\frac{iyz}{\sigma} - \frac{1}{2}y^2\big)\ dy. \\ \textbf{Now it is very solvable with the answer below.} \end{align} Therefore, I first wants to show that \begin{align} p(z) = \frac{1}{\sigma \sqrt{2 \pi}} \mathbb{E}\big[\exp\big(\frac{-(z-X)^2}{2\sigma^2} \big)\big]. \qquad (**) \end{align} To start, \begin{align} \mathbb{E}[e^{iuZ}] &= \mathbb{E}[e^{iu(X+\sigma Y)}]\\ &=\mathbb{E}[e^{iuX}]\mathbb{E}[e^{iu(Z-X)}], \qquad \text{where } (Z-X)\sim \mathcal{N}(0,\sigma^2)\\ (*)&= \mathbb{E}[e^{iuX}]\cdot\frac{1}{\sigma \sqrt{2 \pi}} \int_\mathbb{R} e^{uiz}\cdot e^{\frac{-(z-X)^2}{2\sigma^2}}\ dz. \end{align} However, I am not convinced yet that $(*)$ is valid. And I don't see if this useful for working towards one of the above statements.

Any help is appreciated!

Answer 1

since $X$ and $Y$ are independent we have that $$ E[f(X)\cdot g(Y)] = E[f(X)]\cdot E[g(Y)], $$ for any measurable functions $f,g$ such that the above expressions exist. Since $\sigma Y \sim \mathcal{N}(0,\sigma^2)$ we can use the characteristic function of the normal distribution to obtain the characteristic function of $Z$:

$$ E[\exp(iuZ)] = E\Bigl[\exp\Bigl(iu(X+\sigma Y)\Bigr)\Bigr] = E[\exp(iuX)] \cdot E[\exp(iu\sigma Y)] = \\\phi(u) \cdot \exp\Bigl(-\frac12 u^2\sigma^2\Bigr). $$ Now we can use Fourier Inversion to obtain the density of $Z$:

$$ p(z) = \frac 1{2\pi} \int_{\mathbb{R}} \exp(iuz) \cdot \phi(-u) \cdot \exp\Bigl(-\frac12 u^2\sigma^2\Bigr) \; du. $$ Maybe you can continue from here.

Concerning your equation (*): $$ E[\exp(iu(Z-X))] = E[\exp(iu\sigma Y)] = \frac{1}{\sqrt{2\pi}} \cdot \int_{\mathbb{R}} \exp(iu\sigma y) \exp\Bigl(-\frac12 y^2 \Bigr) \;dy = \\ = \frac{1}{\sqrt{2\pi}\sigma} \cdot \int_{\mathbb{R}} \exp(iuz) \exp\Bigl(-\frac12 \frac{z^2}{\sigma^2} \Bigr) \;dz, $$ where in the last step I used the substituion $z=\sigma\cdot y$.

So your formula is not 100% correct: in particular, $X$ is random variable and should not be the result of an expectation, which is in general a number.

Hope that helps a little.