# Conditional expection of a lognormal process

Suppose I have the following process

$$y_{t+1} = y_t \exp (\varepsilon_{t+1})$$

where $$\varepsilon_{t+1}\sim _{iid} N(\frac{\sigma v^2 }{2},v^2 )$$, $$v> 0$$. I would like to calculate $$E[y_{t+1}|y_t]$$

For this, I apply conditional expectation in the equation above. So

$$E[y_{t+1} | y_t ] =E [y_t \exp (\varepsilon_{t+1})| y_t] = y_t E [ \exp (\varepsilon_{t+1})| y_t].$$

Now I’m unsure of saying that $$\exp (\varepsilon_{t+1})$$ and $$y_t$$ are independent. I suspect this to be true, since $$y_t = y_{t-1} \varepsilon_{t}$$ and $$\{\varepsilon_t\}_{t=0}^{\infty}$$ is a iid process. If the independence of $$\exp (\varepsilon_{t+1})$$ and $$y_t$$ is true, I can conclude that

$$E[y_{t+1} | y_t ] = y_t E [ \exp (\varepsilon_{t+1})| y_t] = y_t E [ \exp (\varepsilon_{t+1})] = y_t \exp( \frac{\sigma v^2 }{2} + \frac{v^2}{2}).$$

Is my argument correct?

• For the conditional expectation with $y_t$ given, there is no need to consider whether $\exp(\epsilon_{t+1}), y_t$ are independent or not. Commented Oct 19, 2020 at 4:31
• I think either i misread your expression or it was changed (earlier it was $y_{t+1}$ I think). Never mind, yes this statement is true. You can use law of total expectations. $$E(\exp(\epsilon_{t+1}))=E(E(\exp(\epsilon_{t+1})|y_t))$$ Commented Oct 19, 2020 at 5:01
• Sorry, but I don't think I can use the law of total expectation to conclude that $E[\exp(\varepsilon_{t+1})| y_t] =E[\exp(\varepsilon_{t+1})]$.
– Fam
Commented Oct 19, 2020 at 5:22
• $E(\exp(\epsilon_{t+1})|y_t) = \exp \bigg( \frac{\sigma v^2 }{2} + \frac{v^2}{2} \bigg)$, which is a constant. Therefore, $$E(E(\exp(\epsilon_{t+1})|y_t))=E\bigg(\exp \bigg( \frac{\sigma v^2 }{2} + \frac{v^2}{2}\bigg)\bigg)= \exp \bigg( \frac{\sigma v^2 }{2} + \frac{v^2}{2} \bigg)$$. Commented Oct 19, 2020 at 6:40
• This is indeed assuming that $y_t$ and $\epsilon_{t+1}$ are independent. Otherwise, there is not enough information in the input. Commented Oct 19, 2020 at 6:43

Your assumption of $$y_t \perp \!\!\! \perp \exp(\varepsilon_{t+1})$$ is correct by definition of $$\varepsilon_t$$, i.e:

$$X \perp \!\!\! \perp Y \Leftrightarrow X \perp \!\!\! \perp f(Y) \ \forall \ f(\cdot) \tag{1}$$

where $$\perp \!\!\! \perp$$ is the independence symbol. In words: if $$X$$ is independent of $$Y$$, then $$X$$ is also independent of any deterministic function of $$Y$$.

Here: $$\varepsilon_t \sim iid(\mu,\sigma^2)$$, i.e. non-zero mean iid noise. Then $$\varepsilon_t$$ is per definition an innovation at time $$t$$, i.e. independent of all variables at $$t$$, i.e. $$y_t, \textbf{X}_t \perp \!\!\! \perp \varepsilon_t$$ and therefore also any deterministic function of it is also independent of all variables at $$t$$, i.e. $$y_t, \textbf{X}_t \perp \!\!\! \perp f(\varepsilon_t)$$ where $$f(x) =exp(x)$$.

For solving $$E[y_{t+1}|y_t]$$ given $$y_{t+1}=y_t \exp(\varepsilon_{t+1})$$:

Option 1:

Take $$E[y_{t+1}|y_t]$$ directly, as you did. The rest is correct.

Option 2:

In cases with exponential functions, it might be more attractive to work with the $$\log$$ of the variables, due to additive properties: Take $$\log(y_{t+1})$$ to get $$\log(y_{t+1})=\log(y_t) + \varepsilon_{t+1}.$$ Then take $$E[\log(y_t)+\varepsilon_{t+1}|y_t]=\log(y_t)+E[\varepsilon_t]=\log(y_t)+\mu$$ Here, independence may be more obvious. However, I can't think of a direct way to get from this to your variable of interest $$E[y_{t+1}|y_t]$$. Note that $$E[\log(y_{t+1})|y_t] \ne \log(E[y_{t+1}|y_t])$$, therefore one can't simply take $$\exp()$$ of $$\log(y_t)+\mu$$.