Black-Sholes Model Related Proof

[DoeeDoe]

I'm given that,
\(S_{t} = S_{0}e^{\nu t+\sigma W(t)}\)

and asked to show that
\(E[S_{t}]= S_{0}e^{\left (\nu + \frac{1}{2}\sigma^{2} \right )t}\)

In these expressions: \(\nu\) is the expected logarithmic return rate. \(\sigma\) is the volatility. and W(t) has a normal distribution with mean 0 and variance t, for each t.

I'm not familiar with taking expectations of a function of a continuous random variable. Therefore I really have no idea how I can prove this. This is a homework question for an introductory financial math class.

Thank you to whoever can help me prove this! I really appreciate you taking the time to help me out with this!

 

[Keith Tan]

ito's lemma! log both sides of the equation:
ln(st)=ln(s0)+ln(e^(ut+sigmaWt))
ln(st)=ln(s0)+(ut+sigmaWt)
d ln(st) = u dt + sigma dWt

by ito's lemma, mean=u and volatilty=sigma.
recall E[e^x] = e^(ut+0.5var(x)), for normally distributed returns of x.

E[St] = S0E[e^(ut+sigmaWt)] = S0e^(uT+0.5var(x)T)

my ito's kind of bad, but you get the drift (or mean).

 

[DoeeDoe]

We haven't actually studied ito's lemma, I kind of follow your process but it is still a little confusing.

I worked out another method (i think), but I'm stuck on this one step.

\(\frac{1}{2\pi t}exp(\frac{\sigma ^{2}t}{2})\int_{-\infty }^{\infty }exp(\frac{-(x-\sigma t)^{2}}{2t}) dx\)

The above expression somehow must equal
\(exp(\frac{\sigma ^{2}t}{2})\)

If I can get it to equal ^, than it would prove the expression. I'm not sure if I the above expressions are incorrect or I just haven't evaluated the integral properly.

 

[horseshoe]

\(E[S_t] = S_0 e^{\nu t} E[e^{\sigma W(t)}]\) From here just note that the expectation is the MGF of a normal RV with mean 0 and variance t, the parameter being \(\sigma\), which is \(e^{1/2 \sigma^2 t}\)

 

[DStahl]

We haven't actually studied ito's lemma, I kind of follow your process but it is still a little confusing.

I worked out another method (i think), but I'm stuck on this one step.

\(\frac{1}{2\pi t}exp(\frac{\sigma ^{2}t}{2})\int_{-\infty }^{\infty }exp(\frac{-(x-\sigma t)^{2}}{2t}) dx\)

The above expression somehow must equal

\(exp(\frac{\sigma ^{2}t}{2})\)

If I can get it to equal ^, than it would prove the expression. I'm not sure if I the above expressions are incorrect or I just haven't evaluated the integral properly.

That integral is equal to one, being the probability density function of a normal random variable. To see this substitute
\( y=\frac{x-\sigma t}{\sqrt{t}} \)

Anyway, the easiest way to solve the problem is as follows:
\( e^ {\sigma W_t} \) solves the SDE
\( dx=x \sigma dW_t+\frac{1}{2} \sigma x dt \)
Integrating, \( x_t=x_0+\frac{\sigma }{2}\int_0 ^ t x ds+ \int_0 ^t \sigma x dW \)
Taking expectations,
\( \mathbb{E}[x_t ]=x_0+\frac{\sigma}{2}\int_0 ^ t \mathbb{E}[x] ds \)

Letting \( y=\mathbb{E} [x] \) and differentiating yields the ODE \(y'=\frac{\sigma}{2} y \)

Which has solution \(y(t)=e^{\frac{\sigma t}{2}} \)

This solves the problem.