1

I am trying to build the mathematical background for an stochastic simulation. Right now I have the mathematical model for:

$$y_t = N(\mu_t, \sigma_t)$$

So that $\mu_t$ and $\sigma_t$ are known. These $y_t$ values are non-independent and are used to evaluate another meassure, let´s say:

$$I_t = (1+y_t) \cdot I_{t-1}$$

Where $I(0)$ is a constant. I am trying to predict the standard deviation of $I_t$ for every $t$.

So far I have tried using the following equations: $$c \cdot N(\mu, \sigma^2) = N(c\mu, (c\sigma)^2)$$ $$c + N(\mu, \sigma^2) = N(c+\mu, \sigma^2)$$ $$N(\mu_1, \sigma_1^2) \cdot N(\mu_2, \sigma_2^2) = N\left(\frac{\sigma_1^2\mu_2+\sigma_2^2\mu_1}{\sigma_1^2+\sigma_2^2}, \frac{1}{\frac{1}{\sigma_1^2}+\frac{1}{\sigma_2^2}}\right)$$

However, after simulating $y_t$ and then converting it to $I_t$ I do not get the same distribution as, with the equations above, I should.

Am I missing something?

EDIT:

I have been able to obtain the mathematical formulation when $\mu_t=0$ $\forall t$ with the help of this link https://mathworld.wolfram.com/NormalProductDistribution.html

To obtain for $\mu_t\neq 0$ I have changed the variables in the integral in the link to: $$P_{XY}(u)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}N_x(0,\sigma_x)N_y(0, \sigma_y)\delta\left((x+\mu_x)(y+\mu_y)-u\right)dxdy$$

How does that affect the Meijer G function the obtain?

  • https://math.stackexchange.com/questions/101062/is-the-product-of-two-gaussian-random-variables-also-a-gaussian – d.k.o. Sep 07 '21 at 15:28
  • So is there a way to evaluate the product of n normally distributed random variables? The post you sent me does it for just 2 variables. – Jorge Rodríguez Peña Sep 08 '21 at 10:27

1 Answers1

0

We have that the process $(I_n)_{n \in \mathbb{N}_0}$ is described by $$I_n=I_0\prod_{k=1}^n(1+y_k)$$ where $y_k \sim \mathcal{N}(\mu_k,\sigma_k^2)$. If we assume that the $y_k$ are independent, the variance is given by $$\textrm{Var}[I_n|I_0]=E[I_n^2|I_0]-E[I_n|I_0]^2$$ therefore $$\begin{aligned}\textrm{Var}[I_n|I_0]&=I_0^2E\bigg[\prod_{k=1}^n(1+y_k)^2\bigg]-I_0^2E\bigg[\prod_{k=1}^n(1+y_k)\bigg]^2=\\ &=I_0^2\bigg(\prod_{k=1}^nE[(1+y_k)^2]-\bigg(\prod_{k=1}^nE[(1+y_k)]\bigg)^2\bigg)=\\ &=I_0^2\bigg(\prod_{k=1}^n(\sigma_k^2+(\mu_k+1)^2)-\prod_{k=1}^n(\mu_k+1)^2\bigg)\\ \end{aligned}$$ $$\implies \textrm{Stdev}[I_n|I_0]=|I_0|\sqrt{\prod_{k=1}^n(\sigma_k^2+(\mu_k+1)^2)-\prod_{k=1}^n(\mu_k+1)^2}$$ If we further assume $\mu_k=\mu$, $\sigma_k=\sigma$ then $$\textrm{Stdev}[I_n|I_0]=|I_0|\sqrt{(\sigma^2+(\mu+1)^2)^n-(\mu+1)^{2n}}$$ And even more conveniently, if $\mu=0$ then $$\textrm{Stdev}[I_n|I_0]=|I_0|\sqrt{(\sigma^2+1)^n-1}$$

Snoop
  • 8,309
  • 3
  • 5
  • 26