Baxter and Rennie

[kimlinski]

On p.42 of the book how has the formula for q under measure Q arrived at from its P equivalent. Thanks for the hint.


On p.137 the equation for B(t) has the integral of W(s) but the equation for P(t,T) starts with product sigma(T-t)*W(t), even though the formula for f(t,T) an r(t) both have sigma*(W(t)). Thanks for the intuition.

Kind regards,

Szczepan

 

[Steve Hales]

Hi
I think I see what you mean. Let me know if I've not understood you correctly.
The first term in the definition of \(P(t,T)\) includes \(\int_{t}^{T}\sigma W_t du\). But notice that there's no \(u\) in the integrand, and so we're essentially integrating a constant. Therefore that integral equals \(\sigma(T-t)W_t\).
Notice that this is not the same as \(\int_{t}^{T}\sigma W_u du\) which depends on the path of the standard Brownian motion rather than its starting value \(W_t\).

 

[kimlinski]

Thanks Steve, integration is done over the maturity of the bond P(t,T) if that makes sense.

How about the formula for the q on p.42 I'm struggling to derive the expression, manipulating the formulae shown before.

 

[Steve Hales]

The risk-neutral up-step probability \(q\) is chosen so that the share price is expected to grow that the risk-free rate over each time-step. Algebraically this mean:
\(q\times s_{up} + (1-q)\times s_{down} = s \times exp(r\delta)\)
Rearranging will result in the page 42 definition for \(q\).

 

[kimlinski]

Thanks Steve. Should've been more precise in the original question: p.42 has two definitions of the q and itsv the second that gives me headaches. Precisely, it's
q=(1/2)*(1-((delta*t)^(1/2))*[mju+(1/2)*sigma^2-r]/sigma)
Thanks in advance

 

[Steve Hales]

OK, I see what you mean.
Start with the definition for \(q\), and replace \(s_{up}\) and \(s_{down}\) with their respective values found on page 41, which leads to:

\[
q=\frac{exp(r\delta t)-exp(\mu\delta t - \sigma\sqrt{\delta t})}{exp(\mu\delta t + \sigma\sqrt{\delta t}) - exp(\mu\delta t - \sigma\sqrt{\delta t})}
\]
Now use Taylor's approximation in each of the four exponential functions: \(exp(x) \approx 1 + x +\tfrac{1}{2}x^2\).
This then simplifies (remember that powers of \(\delta t\) higher than 1 go to zero) to the second expression for \(q\) on page 42.

 

[kimlinski]

Sorry to resurrect an old thread, but I have some doubts about the wording on page 42.
The formula in paragraph :

"Thus [2Xn-n]/[n]^(1/2) has mean -[t]^(1/2)×[[mu+1/2×sigma^2-r]/sigma]...

The [2Xn-n]/[n]^(1/2) is the product of multiplication of the [Xn-[n/2]] / [n/4]^[1/2]

Where probability is equal to 1/2 so that nq is n/2 and nq(1-q) is n/4, therefore the [Xn-nq]/[nq[1-q] when probability is

q=[s[exp(rt)]-s_down]/[s_up-s_down] ( before polynomial expansion ) does not equal

[2Xn-n]/[n]^(1/2)

Therefore the [2Xn-n]/[n]^(1/2) would have to be also the result of the formula [Xn-nq]/[nq[1-q]]^(1/2)], where probability q=[s[exp(rt)]-s_down] / [s_up-s_down] after expansion is equal to [r×del_t - mu×del_t + [sigma×[del_t]^(1/2)-[[sigma^2]/2]×del_t] / [2×sigma×(del_t)^(1/2)

Thanks in advance while I continue my investigation.

 

[kimlinski]

Now, I was able to arrive at the result that the variance nq(1-q) is equal to n/4, when q after polynomial series expansion is
[r*del_t - mu*del_t + sigma*del_t +(sigma^2*del_t)/2] / 2 when terms del_t of power higher than 1 are ignored, but I'm struggling to arrive at the distribution of St to be intuitively convincing as N(r*t - (sigma^2)/2*t; sigma^2*t), when applying the q to formula (Xn-nq)/(nq(1-q))^(1/2), going back to the drawing board.