Sampling from the product of Normal and Log-normal distributions Here is my problem:

*

*Let us generate random samples $X$ from Normal distribution with $\mu=0$ and $\sigma^2 =0.046$.


*Then, using the following equations, the corresponding random samples $Y$ generated from Log-Normal distribution will have $\mu_{LN}= 1.023$ and $\sigma_{LN}^2=0.222$.
\begin{equation}
        \begin{aligned}
        \mu_{LN} &= \mathrm{e}^{(\mu + \frac{\sigma^{2}}{2})} \\
        \sigma_{LN}^{2} &= \mathrm{e}^{(2 \mu + \sigma^{2})} \left(\mathrm{e}^{\sigma^{2}}-1\right)
    \end{aligned}
    \end{equation}


*Afterwards, I drew the PDF and histogram of $X\sim \mathcal{N}(\mu, \sigma^2) = \mathcal{N}(0, 0.046)$ and $Y\sim \mathcal{LogN}(\mu_{LN}, \sigma^2_{LN}) = \mathcal{LogN}(1.023, 0.222)$, which are given in the following figures: Fig: Normal and Fig: LogNormal


*Let us assume that $Z=XY$. The histogram of the product (i.e, $Z$) is shown in the following figure Fig: Hybrid.
Here are the questions I am addressing:

*

*How can I calculate the mean $\mu_Z$ and variance $\sigma^2_{Z}$ of $Z$? I have tried the mathematical formulations given in this Link. Given $\mu=0$ and $\sigma^2 =0.046$, then $\mu_Z$ and $\sigma^2_{Z}$ will be 0 and 0.05, respectively. However, the histogram of the product $Z$ has $\sigma^2_{Z} \approx 0.5$ instead of 0.05. I guess the reason behind the difference in $\sigma^2_{Z}$ is due to the fact that $X$ and $Y$ are correlated in our case, am I right?
If so, what is the true value of $\mu_Z$ and $\sigma^2_{Z}$, in our case? How can I calculate the PDF of $Z$?

*Assuming that $\mu=0$ and $\sigma^2$ has a very small value, can we consider PDF of the product $Z$ as an approximately Normal distribution? To verify this, I calculated the mean and variance of the samples generated from $Z$, which are: $\mu_Z = 0$ and $\sigma^2_{Z} = 0.5$. Then, I drew the PDF of $Z\sim \mathcal{N}(0, 0.5)$ on the same figure with the histogram of $Z$ (See this Figure).

Finally, I would be grateful if you can recommend some articles or books for tackling this issue.
Many thanks in advance!
 A: First of all notice, that the mean and variance of a $\mathcal{LogN}(\mu,\sigma^2)$ distribution are given by
$$\mu_{LN} = e^{\mu + \sigma^2/2} \quad \text{ and } \quad \sigma_{LN}^2=e^{2\mu +\sigma^2}(e^{\sigma^2} - 1).$$
But this means that the mean and variance of $Y\sim \mathcal{LogN}(\mu_{LN},\sigma_{LN}^2)$ is not $\mu_{LN}$ and $\sigma^2_{LN}$ but rather
$$\mathbb{E}[Y] = e^{\mu_{LN} + \sigma_{LN}^2/2} \quad \text{ and } \quad \operatorname{Var}(Y)=e^{2\mu_{LN} +\sigma_{LN}^2}(e^{\sigma_{LN}^2} - 1).$$

Using the formula in your mentioned link you do get, that the variance of $Z=XY$ is actually
\begin{align*}
\operatorname{Var}(XY) &=(\mu^2+\sigma^2)\exp\left(2\mu_N+2\sigma_N^2\right)-\mu^2\exp\left(2\mu_N+\sigma_N^2\right) \\
&= (0^2 + 0.046)\exp(2\cdot 1.023 + 2\cdot 0.222) \approx 0.555
\end{align*}
However if you replace $Y$ with $Y\sim \mathcal{LogN}(\mu,\sigma^2)$, then you get
\begin{align*}
\operatorname{Var}(XY) = (0.046)\exp(2\cdot 0.046) \approx 0.05,
\end{align*}
which may be the source of your error.

Regarding the PDF there is a general formula for calculating the PDF for a product of two independent random variables. In your case the PDF becomes
\begin{align*}
f_Z(z)=\frac{1}{2\pi}\frac{1}{\sigma \cdot \sigma_{LN}} \int_0^\infty \exp(-\frac{z^2}{2\sigma^2 x^2}) \exp(-\frac{(\ln(x)-\mu_{LN})^2}{2\sigma_{LN}^2}) \frac{1}{x^2} \: dx,
\end{align*}
which may be hard to compute analytically, however it is worth noting that $f_Z(z)=f_Z(-z)$, so the distribution is symmetric around $0$, which is also what your histogram suggests.

Regarding asymptotic normality of $Z$ as $\sigma^2$ goes to $0$, i do believe that this is the case. I will not give the full argument here, but rather a sketch of a proof. The first thing to note is that the distribution of $Y$ can be written as $Y=e^W$ where $W\sim N(\mu_{LN},\sigma^2_{LN})$, furthermore note, that
$$\lim_{\sigma^2 \rightarrow 0} \frac{\sigma^2_{LN}}{\sigma^2}=e^{2\mu }\lim_{\sigma^2 \rightarrow 0}\frac{(e^{\sigma^2} - 1)}{\sigma^2} =e^{2\mu}=1$$
assuming $\mu=0$. We now have, that $Z=Xe^W$ can be written as a smooth function of two independent normal distributions with variance tending to $0$ as $\sigma^2 \rightarrow 0$. It can thus be shown using the delta method that the distribution of $Z$ is asymptotically normal.
