2013 Sept 5i

[Jia Syuen]

May I know how we get the variance parameter for S_t/S_0 as 0.25t? Thank you in advance.

 

[Steve Hales]

Hi
The solution to the SDE is:

\(S_t = S_0 e^{0.275t + 0.5B_t}\)

Therefore:

\(ln(S_t/S_0) = 0.275t + 0.5B_t\)

From the definition of standard Brownian motion, \(B_t \sim N(0,t)\), therefore \(0.5B_t \sim N(0,0.25t)\) and:

\(ln(S_t/S_0) \sim N( 0.275t ,0.25t)\)

Steve

 

[Jonathan]

Hi Steve

Can you please explain how you get to 0.5Bt~N(0,0.25t) from Bt~N(0,t)?

Can you also please explain how you get the mu parameter for the distribution of St?

Thanks

 

[Steve Hales]

Since Bt~N(0,t) then 0.5Bt will be normally distributed, we just need to know its mean and variance:

• E[0.5Bt]=0.5*E[Bt]=0
• Var(0.5Bt) = 0.5^2 * Var(Bt) = 0.25t

Therefore 0.5Bt~N(0,0.25t).
Hope that helps.

 

[Jonathan]

Thanks Steve that is really helpful. I knew that I was missing something simple...

Can you explain how, in this question, we get to S_t ~ logN(logS_0 + 0.275t, 0.25t)?

I understand the variance part as you have explained above.

Thanks

 

[Steve Hales]

Hi
If you're happy with \(ln(S_t/S_0) \sim N( 0.275t ,0.25t)\) from above, then we just need to note that:

\(ln(S_t/S_0) = ln(S_t) - ln(S_0)\)

Therefore we have:

\(ln(S_t) - ln(S_0) \sim N( 0.275t ,0.25t)\)

which implies that:

\(ln(S_t) \sim N(ln(S_0)+ 0.275t ,0.25t)\)

Let me know if that's not what you meant.