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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Dayne
    Oct 19, 2020 at 4:31
  • 1
    $\begingroup$ 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. $$E(\exp(\epsilon_{t+1}))=E(E(\exp(\epsilon_{t+1})|y_t))$$ $\endgroup$
    – Dayne
    Oct 19, 2020 at 5:01
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    $\begingroup$ 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})] $. $\endgroup$
    – Fam
    Oct 19, 2020 at 5:22
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    $\begingroup$ $E(\exp(\epsilon_{t+1})|y_t) = \exp \bigg( \frac{\sigma v^2 }{2} + \frac{v^2}{2} \bigg)$, which is a constant. Therefore, $$E(E(\exp(\epsilon_{t+1})|y_t))=E\bigg(\exp \bigg( \frac{\sigma v^2 }{2} + \frac{v^2}{2}\bigg)\bigg)= \exp \bigg( \frac{\sigma v^2 }{2} + \frac{v^2}{2} \bigg)$$. $\endgroup$
    – Dayne
    Oct 19, 2020 at 6:40
  • 2
    $\begingroup$ This is indeed assuming that $y_t$ and $\epsilon_{t+1}$ are independent. Otherwise, there is not enough information in the input. $\endgroup$
    – Xi'an
    Oct 19, 2020 at 6:43

1 Answer 1


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 $\log(y_t)+\mu$.


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