Is the value also log-normally distributed?

My book assumes many times that $log(1+R)$ is normally distributed, so R is log-normal. But does this also mean that the value process is log-normal? Since $V=V_0(1+R)\rightarrow V/V_0=1+R$, and since $1+R$ is log-normal, $V/V_0$ is log-normal, so $log(V/V_0)=log(V)-log(V_0)$ is normal, and hence V is also log-normally distributed?

1 Answer

There are many ways answering this, here is one:

We assume the asset price at $t=T$, $S_T = S_{T-1} \times (S_T / S_{T-1})$.

Assuming continuous compounding, we can write, $S_T = S_{T-1} \times \exp(R_{T-1})$.

Working the same way for the previous period, we get $S_{T} = S_{T-2} \times \exp(R_{T-1}+R_T)$.

Working all the way back to the initial value of the asset price, $S_0$, we have that: $S_T = S_0 \times \exp(R_1 + R_2 + ... + R_T)$.

Dividing by $S_0$ and taking $\log$s yields: $\ln(S_T/S_0) = R_1 + R_2 + ... + R_T$.

Here, $(S_T/S_0)$, which is the return of the asset over the entire period, is lognormally distributed if $R_1 + R_2 + ... + R_T$ is normally distributed. To arrive to the appropriate format of the lognormal distribution,and since $R$ is normally distributed, by $\exp$ing: $(S_T/S_0) = \exp(R_1 + R_2 + ... + R_T)$

Hence, to pose on another way, if we assume that prices are lognormally distributed (which could not always be the case), then $\log(1+R_t)$ is lognormally distributed.