In finance, the price of an asset is given by the following formula.

$P_t=P_0* e^r$

Pt = Price at time "t".
P0 = Actual Price.

Likewise, it is assumed that the yields (r) will follow a normal distribution with mean $u$ and variance $\sigma^2$.

so, the distribution functions $$ F_p(p) = P(p_t\le p_t)=P(p_0 *e^r\le p_t)=P(e^r\le \frac{pt}{p_0}) =P(r<=\ln(\frac{P_t}{P_0})) = F_r(r=\ln(\frac{P_t}{P_0})) $$ To obtain the density function and using the chain rule, we derive the distribution functions,

$$ f_p(p)=f_r(\ln(\frac{P_t}{P_0}))*\frac{1}{P_t}$$ By hypothesis we have that the density function of r follows a normal with zero mean and variance sigma squared, then

$$=>\frac{1}{P_t\sqrt{2\pi \sigma^2}}\exp\{-(\ln(\frac{P_t}{P_0})^2/(2\sigma^2)\}$$ Applying the logarithm property, we have

$$=>\frac{1}{P_t\sqrt{2\pi \sigma^2}}\exp-\{(\ln(P_t)-\ln(P_0))^2/(2\sigma^2)\}, $$ Concluding that , $P_t \sim \text{LogNormal}(\ln(P_0), \sigma^2)$.

Is this derivation of price theory in finance correct? That is, prices are distributed lognormally with mean $\ln(P_o)$ and variance $(\sigma^2)$.

I ask the question because a log-normal distribution has mean = $\exp^(u+\frac{\sigma^2}{2})$ and not $\ln(x_0)$.

Please someone who can support me.


1 Answer 1


I am assuming you are referring to stocks as financial assets instead of derivatives. Sometimes, stock price behavior is said to follow geometric Brownian motion. You can look up if unfamiliar. If you follow that assumption and plug a function G = ln(Stock price) into something called Ito's lemma, you can derive a distribution kind of similar to what you have in the last sentence of your answer: ln(St) ~ N[lnS0 + (u - σ^2/2)T, σ^2T]

From here you can derive what the distribution of St will look like instead of ln(St). I am not really sure what you did in your proof sorry.

  • $\begingroup$ @usuario357057 Yes, I know that the prices of financial assets are modeled with the GBM. But imagine I have a model $P_t = P_0 * e^x$ (also known as log-lin model ). So what distribution does Pt have if I am aware that "r" is normally distributed $(r \sim N (0,\sigma^2)$. According to the steps described above, Pt is lognormally distributed. $P_t \sim LogNorm(Ln(P_0),\sigma^2)$. I wanted to know if this deduction is correct. Thank you very much. $\endgroup$
    – Christian
    Commented May 1, 2022 at 22:24

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