# Expected value of log-GARCH process

Is there a way to analitycally compute expectation of log-GARCH process?

The GARCH(1,1) process: $dU_t = \theta(\omega - U_t) dt + \xi U_t d W_t$

The log-GARCH(1,1) process: $e^{U_t}$

The expectation I'm interested in: $m(t) = E[e^{U_t}]$

I can't find any explicit form (or any) of the GARCH process distribution, not to mention log-GARCH. Nethertheless, I have simulated $E[e^{U_t}]$ and it seems exponential, so I suspect, there migth be an explicit formula for $E[e^{U_t}]$. Is there?

• Why is this a Garch process in the first place? – Kiwiakos Nov 1 '15 at 4:24
• @Kiwiakos: I think, this is because $U_t$'s power is 1 (as it is in what's traditionally called GARCH(1,1)) in $\xi U_tdW_t$, as opposed to other models like square-root and $\frac{3}{2}$. – Oleg Nov 1 '15 at 10:41

To solve for $U_t$, we can proceed as follows. First, note that \begin{align*} d\left(e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t \right) &= e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t \left((\theta+\xi^2) dt -\xi dW_t\right) \\ &\qquad+ e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} dU_t -\xi^2e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} U_t dt\\ &=\theta \omega e^{(\theta + \frac{1}{2}\xi^2)t - \xi W_t} dt. \end{align*} Then \begin{align*} U_t &= U_0 e^{-(\theta + \frac{1}{2}\xi^2)t + \xi W_t } + \theta \omega \int_0^t e^{-(\theta + \frac{1}{2}\xi^2)(t-s) + \xi (W_t-W_s)} ds. \end{align*} From here, we can compute $E(U_t)$ analytically.

However, for Expectation $E(e^{U_t})$, we note the following. Let $\eta$ be a standard normal random variable. Then \begin{align*} E\left(e^{(e^{\eta})} \right) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{(e^x-1/2x^2)}dx \\ &=\infty, \end{align*} as \begin{align*} \lim_{x \rightarrow \infty}e^{(e^x-1/2x^2)} = \infty. \end{align*}

Now, we consider $E\left(e^{U_t}\right)$. Note that, for $\theta \omega \geq 0$ and $U_0>0$, \begin{align*} U_t &= U_0 e^{-(\theta + \frac{1}{2}\xi^2)t + \xi W_t } + \theta \omega \int_0^t e^{-(\theta + \frac{1}{2}\xi^2)(t-s) + \xi (W_t-W_s)} ds\\ &\geq U_0 e^{-(\theta + \frac{1}{2}\xi^2)t + \xi W_t }. \end{align*} Then, \begin{align*} E(e^{U_t}) &\geq E\left(e^{U_0 e^{-(\theta + \frac{1}{2}\xi^2)t + \xi W_t }} \right)\\ &=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{U_0 e^{-(\theta + \frac{1}{2}\xi^2)t + \xi \sqrt{t} x-\frac{1}{2}x^2}}dx\\ &= \infty. \end{align*}

NOTE: If $\omega =0$, then $U_t$ is log-normal. Related information: In the interest rate world, if the short rate $r_t$ is log-normal, then the money market account value $B_t = e^{\int_0^t r_s ds}$ has infinite expectation; see Page 63 of the book Interest Rate Models.

• Thank for the answer! Though I have a doubt concerning the first line: it seem that you have applied the differential product rule here - $d(uv) = udv + vdu$. But while in usual calculus $dvdu$ is infinitesimalvand doesn't count, in stochastics it yields additional term due to $dW_t dW_t = dt$. – Oleg Oct 30 '15 at 6:50
• Here it will be: $$d e^{(\theta - \frac{1}{2}\xi^2)t - \xi W_t} dU_t = - \xi^2 U_t e^{(\theta - \frac{1}{2}\xi^2)t - \xi W_t} dt$$ So the second line will turn to be integral equation and i'm not sure the following analysis will still be relevant. – Oleg Oct 30 '15 at 6:56
• @Oleg: you are right. I will make the correction. – Gordon Oct 30 '15 at 12:33
• @Oleg: Thanks for your comments. I have made the correction by changing $\theta-\frac{1}{2}\xi^2$ to $\theta+\frac{1}{2}\xi^2$. The remaining analysis does not change. – Gordon Oct 30 '15 at 12:45
• Nice technique, thanks! But I'm still not sure that the $E(e^{e^\nu})$-argument proves that $E(e^{U_t})$ doesn't exist. Because, you know, $U_t$ has that integtal term apart from log-normal term. Besides, I've made a simulation of $E(e^{U_t})$ and got some finite result for all $t$. Moreover, the whole problem comes from computing expected volatility of log-normal stochastic volatility in SV-model with GARCH-diffusion process for volatility. Namely, $E(e^{U_t})$ - is that expected volatility. Thus, I expect that expectation to exist, otherwise it would be a useless model. – Oleg Oct 31 '15 at 5:59