# Log of square of Geometric Brownian Motion

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

• I'd say the first one is wrong because $dF^2 / F^2 \neq d\ln F^2$. You are shoving some second order terms under the rug in this step. – Raskolnikov Apr 25 '19 at 7:05
• Thanks for confirming, I also thought this step had an issue. However, the <b>Second</b> also has some other issue, since the answer I get is wrong. The correct answer (as per various literature) is that the sign should be "+" for the $\sigma^2 dt$ term – bhutes Apr 25 '19 at 7:08

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

• Thanks. Now I understand it. – bhutes Apr 26 '19 at 0:10

The first one is wrong.

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

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

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

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.

• Thanks. This calculation improved my understanding by a huge amount. – bhutes Apr 25 '19 at 7:42

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.

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.

• Agree. The answer I get in the Second is wrong (some other issue, I think). The link quant.stackexchange.com/questions/31950/… says the $\sigma^2 dt$ should have a "+" sign. – bhutes Apr 25 '19 at 7:13
• I don't see anything in that link that indicates there should be a "+" sign. – Raskolnikov Apr 25 '19 at 7:17
• From the result of $d \ln F^2$ could you obtain the result for $E(F^2)$. In the literature, it says it should be $F(0)^2e^{\sigma^2 T}$. Note, it is not $e^{-\sigma^2 T}$ – bhutes Apr 25 '19 at 7:23
• Yes, but note $T$ is also the terminal time. Who's to say that there's no cancelation of signs going on here? Can you show more explicitly what you are aiming to do? – Raskolnikov Apr 25 '19 at 7:33
• 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 $E(F(T)^2)=F(0)^2 e^{\sigma^2 T}$. – bhutes Apr 25 '19 at 7:38