Log of square of Geometric Brownian Motion

Question

Which of the two calculations below, is wrong? Why?

$dF = \sigma F dW$

First:

$dF^2 = (F^2)' dF + \frac{1}{2}(F^2)''dF.dF$

$dF^2 = 2F dF + dF.dF$

$dF^2 = 2 \sigma F^2 dW + \sigma^2 F^2 dt$

$\frac{dF^2}{F^2}=2\sigma dW + \sigma^2 dt$

$d \ln F^2=2\sigma dW + \sigma^2 dt$

OR

Second:

$d \ln F^2 = (\ln F^2)' dF + \frac{1}{2}(\ln F^2)'' dF.dF$

$d \ln F^2 = \frac{2F}{F^2} dF + \frac{1}{2} (\frac{2F}{F^2})' dF.dF$

$d \ln F^2 = \frac{2}{F} dF +\frac{1}{2} (\frac{2}{F})' dF.dF$

$d \ln F^2 = \frac{2}{F} dF +\frac{1}{2} \frac{-2}{F^2} dF.dF$

$d \ln F^2 = \frac{2}{F} dF - \frac{1}{F^2} dF.dF$

$d \ln F^2 = 2\sigma dW - \sigma^2 dt$

---

Just for context, I am trying to understand the calculation of In-Arears Swap pricing. So, need to compute expectation of "Square of Forward rate".

The book (Brigo Mercurio) says $E(F(T)^2)=F(0)^2 e^{\sigma^2 T}$

My computations above (Second) tells me that $E(F(T)^2)=F(0)^2 e^{-\sigma^2 T}$

I am stuck and unable to explain the "-" sign before $\sigma^2 dt$ which I see.

Ref: Brigo and Mercurio "Interest Rates Models - Theory and Practice" Second Edition, Part V, Chapter 13, Equation 13.3

Answer 1

The second solution is correct, so the solution at t is:

$F_t^2=F_0^2e^{2 \sigma W_t-\sigma^2t}$

Now apply expectation to both sides.

$ E\left[F_t^2\right]=F_0^2 \, E\left[e^{2 \sigma W_t-\sigma^2t}\right]$

The term in the exponent is just Gaussian so let’s call it X: $X=2 \sigma W_t-\sigma^2t$

It’s mean and variance are:

$E[X]=-\sigma^2t$

$V[X]=4 \sigma^2t$

And hence: $E[X]+\frac{1}{2}V[X]=\sigma^2t$

And therefore: $ E\left[F_t^2\right]=F_0^2 \, E\left[e^{X}\right] =F_0^2 e^{\sigma^2t}$

For the mean and half the variance business above, please see the discussion here:https://math.stackexchange.com/questions/176196/calculate-the-expected-value-of-y-ex-where-x-sim-n-mu-sigma2

Answer 2

The first one is wrong.

As I said in my comment, you seem to be shoving some terms under the rug. In fact:

$$d\ln F^2 = \frac{1}{F^2}dF^2 + \frac{1}{2}(-\frac{1}{F^4})dF^2\cdot dF^2 \; .$$

Which leads to

$$d\ln F^2 = \frac{1}{F^2}dF^2 - 2\sigma^2dt \; .$$

Hence with this correct substitution in the first computation, both computations will lead to the same result.

By the way, $\ln F^2 = 2\ln F$. This gives an even more straightforward confirmation that this is the correct formula.

Answer 3

I just want to offer an alternative approach to the problem. I simply used Ito's Lemma. We have $$ dF_t = \sigma F_t dW_t.$$ Now, we want to know the dynamics of $\ln F_t^2$ (https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma). We set $f(t, x) = \ln x^2$ and get the derivatives $$ \frac{\partial f(t, x) }{\partial t} = 0, \quad \frac{\partial f(t, x)}{\partial x} = \frac{2}{x}, \quad \frac{\partial^2 f(t, x)}{\partial x^2 }= -\frac{2}{x^2}$$ It follows \begin{align*} df(t,F_t) & = \frac{\partial f(t, x) }{\partial t}dt + \frac{\partial f(t, x)}{\partial x} dF_t + \frac{1}{2} \frac{\partial^2 f(t, x)}{\partial x^2 }(dF_t)^2\\ & = 0 + \frac{2}{F_t}\sigma F_t dW_t + \frac{1}{2}\frac{-2}{F_t^2}\sigma^2F_t^2dt\\ & = 2\sigma dW_t -\sigma^2dt \end{align*}

You probabably did it the same way as I. However, this is also to show an easy application of Ito's Lemma to a maybe not so sophisticated reader.

Answer 4

the second one is correct because you have used the right tools to compute $d \ln F^2$.

In the first one everything up to the last conclusion is correct.

However,

$$ \frac{dF^2}{F^2} = 2\sigma dW + \sigma^2dt $$

does not imply $$ d \ln F^2 = 2\sigma dW + \sigma^2 dt. $$

This is only true if $F$ would not be raised to the power of two.