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I am looking for help in understanding the algebraic derivation to go in between some of the lines in Pat Hagan's famous Convexity Conundrums paper e.g. how he goes from 3.4a to 3.5a.

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    $\begingroup$ I think this is a legitimate question. However, you are asking people to read the paper and answer your question, which appears time consuming. It will be more helpful that you write out the steps and formulas and state where you do not understand. $\endgroup$
    – Gordon
    Nov 24, 2015 at 16:36
  • $\begingroup$ i agree. but for my q, the only way to answer is to read the whole paper, as without doing so, 1 might miss something importat to the solution (3.4a - 3.5a are right at the conclusion of the paper) $\endgroup$
    – Randor
    Nov 24, 2015 at 16:41
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    $\begingroup$ A question like this is certainly on-topic here. $\endgroup$
    – Bob Jansen
    Nov 24, 2015 at 20:29

1 Answer 1

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Swap

Just to be clear, (3.4c) leads to (3.5a) when we assume lognormal $R(\tau)$. Lognormal $R(\tau)$ means we can write

$$R(\tau) = R_0 e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z}$$

with $Z$ normal, and I'm assuming a zero mean -- which I think is required. Then for (3.4c) we have for the expectation value:

$$ E\left[(R(\tau) - R_0)^2 \right] = (R_0)^2 E\left[(e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z} - 1)^2 \right] = (R_0)^2 E\left[e^{-\sigma^2 \tau + 2 \sigma \sqrt{\tau} Z} - 2 e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z} + 1\right] $$ Now, recall that: $$E[e^{a Z}] =\int_{-\infty}^{\infty} e^{a Z} \frac{e^{-Z^2/2}}{\sqrt{2\pi}} dZ = e^{\frac{a^2}{2} }\int_{-\infty}^{\infty} \frac{e^{-(Z-a)^2/2}}{\sqrt{2\pi}} dZ = e^{\frac{a^2}{2} } $$ So $$ (R_0)^2 E\left[e^{-\sigma^2 \tau + 2 \sigma \sqrt{\tau} Z} - 2 e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z} + 1\right] = (R_0)^2 (e^{-\sigma^2 \tau +2\sigma^2\tau }- 2 e^{-\frac{1}{2}\sigma^2 \tau +\frac{1}{2}\sigma^2\tau} +1)=(R_0)^2 (e^{\sigma^2 \tau}-1)$$

So this explains how (3.5a) follows from (3.4c) (you can fill in the remaining factors yourself).

Cap

Next I'll look at (3.4a) which, after assuming lognormality of $R$, results in (3.5b). The expectation value of interest is

$$E[(R(\tau) − RR_0)[R(\tau) − K]^+] = E[R(\tau) [R(\tau) − K]^+] - R_0 E[[R(\tau) − K]^+]$$

The second term is the usual expression for a call option, so in the Black scholes framework this is simply given by

$$R_0 E[[R(\tau) − K]^+] =R_0(R_0 \mathcal{N}(d_{1/2}) - K\mathcal{N}(d_{-1/2}))$$

You can already spot these terms in expression (3.5b). The other expectation value is a bit trickier, as we have

$$E[R(\tau) [R(\tau) − K]^+] = R_0 E[e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z} [R_0e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z} − K]^+] $$

This is of the form $$ E[ S e^{aZ} [S e^aZ - K]^+] = S\int_{-\infty}^{\infty} e^{a Z} [S e^{aZ} - K]^+\frac{e^{-Z^2/2}}{\sqrt{2\pi}} dZ $$ $$=Se^{\frac{a^2}{2}}\int_{-\infty}^{\infty} [S e^{aZ} - K]^+\frac{e^{-(Z-a)^2/2}}{\sqrt{2\pi}} dZ $$

Next we sub $Z\rightarrow W = Z-a$, giving

$$Se^{\frac{a^2}{2}}\int_{-\infty}^{\infty} [S e^{a(W+a)} - K]^+\frac{e^{-W^2/2}}{\sqrt{2\pi}} dW = Se^{\frac{a^2}{2}} E[S e^{a^2 + aW} - K]^+]$$

Now recall that $a=\sigma\sqrt{\tau}$ and $S=R_0 e^{-\frac{1}{2}\sigma^2 \tau}$. Then this expectation value becomes

$$ R_0 E[R_0 e^{\frac{1}{2}\sigma^2 \tau + aW} - K]^+] = R_0 E[e^{\sigma^2 \tau} R(\tau) - K]^+]$$

That was a lot of effort just to show that:

$$E[R(\tau) [R(\tau) − K]^+] = R_0 E[e^{\sigma^2 \tau} R(\tau) - K]^+]$$

provided $R(\tau)$ is lognormal with zero mean. The right hand side is just a call option, but with spot equal to $R_0 e^{\sigma^2 \tau}$. This gives

$$E[R(\tau) [R(\tau) − K]^+] = (R_0)^2 e^{\sigma^2 \tau}\mathcal{N}(e_{1/2}) - R_0 K\mathcal{N}(e_{-1/2})$$

where $$ e_{1/2} = \frac{\log(\frac{R_0 e^{\sigma^2 \tau}}{K}) + \frac{1}{2}\sigma^2 \tau}{\sigma \sqrt{\tau}} =\frac{\log(\frac{R_0}{K}) +\frac{3}{2}\sigma^2 \tau}{\sigma \sqrt\tau} = d_{3/2}$$ $$ e_{-1/2}=\frac{\log(\frac{R_0 e^{\sigma^2 \tau}}{K}) -\frac{1}{2}\sigma^2 \tau}{\sigma \sqrt{\tau}} =\frac{\log(\frac{R_0}{K}) +\frac{1}{2}\sigma^2 \tau}{\sigma \sqrt\tau} = d_{1/2}$$

So:

$$E[R(\tau) [R(\tau) − K]^+] = (R_0)^2 e^{\sigma^2 \tau}\mathcal{N}(d_{3/2}) - R_0 K\mathcal{N}(d_{1/2})$$

Finally,

$$\begin{equation} E[(R(\tau) − R_0)[R(\tau) − K]^+]= (R_0)^2 e^{\sigma^2 \tau}\mathcal{N}(d_{3/2}) - R_0 K\mathcal{N}(d_{1/2}) - R_0(R_0 \mathcal{N}(d_{1/2}) - K\mathcal{N}(d_{-1/2})) \\ = (R_0)^2 e^{\sigma^2 \tau}\mathcal{N}(d_{3/2}) - R_0(R_0+K)\mathcal{N}(d_{1/2}) + R_0 K\mathcal{N}(d_{-1/2}) \end{equation}$$

This is (part of) equation (3.5b). This should convince you that (3.5b) follows from (3.4a).

Floor

I'll leave you this one.

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  • $\begingroup$ thank you very much Olaf for your detailed answer. this is exactly what i was looking for. very much appreciated :) also, to the moderators, i am very impressed with this site :) $\endgroup$
    – Randor
    Nov 25, 2015 at 20:41

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