Consider the following model for $Y_t$:

$\Delta$log($Y_{T+1})$ = $u_T$ where $u_T$ ~ IID Normal(0,$\sigma^2$).

I want to forecast $Y_{T+1}$. Taking exponentials and then expectations, we see that

$E(Y_{T+1}|\Omega_T)$ = $Y_T E(e^{u_{T+1}}|\Omega_T)$ = $Y_TE(e^{u_{T+1}})$ since $u_{T+1}$ is IID. Also, since $u_t$ ~ Normal(0,$\sigma^2$), we know (from the moment generating function) that $E(e^{u_t})$ = $e^{(1/2)\sigma^2}$

Question: Why is there a tendancy for the one step ahead forecast to be above its previous value, and not below. In mathematical terms, why is $E(Y_{T+1}|\Omega_T) \geq Y_T$?


1 Answer 1


You may be thinking:

"My specification is $Y_{T+1} = Y_Te^{u_{T+1}}$. Since $u_{T+1}$ takes symmetrically positive and negative values, we have that $e^{u_{T+1}}$ can be higher or lower than unity. So $Y_{T+1}$ can be higher or lower than $Y_T$. Why then the conditional expected value is $E(Y_{T+1}|\Omega_T) \geq Y_T$?"

The answer is "because exponentiation is not a symmetry-preserving transformation". While a normal rv has its probability mass allocated symmetrically around its mean/median/mode, this does not happen with its exponentiated transformation, i.e. with a log-normal random variable, which has a mean value always greater than unity. So, "on average", the one-step-ahead forecast will be above the previous value of this specific process, which is by construction a sub-martingale.


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