Q:

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

A:

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

A:

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.

