Spotting the faulty step in two Ito derivations

Question

Let $dM=\sigma M\,dW_t$. Two attempts at finding $d\ln M^2$ are shown below; they disagree, so one contains an error. Put $a=1$ if the mistake is in Method 1 and $a=2$ if it is in Method 2, and let $l$ be the number of the first wrong line in that method. Give $100a+l$. Method 1 $ \begin{aligned} &(1)\quad dM^2=(M^2)'\,dM+	frac12(M^2)''\,d[M,M]\ &(2)\quad dM^2=2M\,dM+d[M,M]\ &(3)\quad dM^2=2\sigma M^2\,dW_t+\sigma^2M^2\,dt\ &(4)\quad \frac{dM^2}{M^2}=2\sigma\,dW_t+\sigma^2\,dt\ &(5)\quad d\ln M^2=2\sigma\,dW_t+\sigma^2\,dt \end{aligned} $ Method 2 $ \begin{aligned} &(1)\quad d\ln M^2=(\ln M^2)'\,dM+	frac12(\ln M^2)''\,d[M,M]\ &(2)\quad d\ln M^2=\frac{2M}{M^2}\,dM+	frac12\left(\frac{2M}{M^2}ight)'d[M,M]\ &(3)\quad d\ln M^2=\frac{2}{M}\,dM+	frac12\left(\frac{2}{M}ight)'d[M,M]\ &(4)\quad d\ln M^2=\frac{2}{M}\,dM-	frac12\cdot\frac{2}{M^2}\,d[M,M]\ &(5)\quad d\ln M^2=\frac{2}{M}\,dM-\frac{1}{M^2}\,d[M,M]\ &(6)\quad d\ln M^2=2\sigma\,dW-\sigma^2\,dt \end{aligned} $

Accepted Answer

$105$. Method 1 breaks at line 5: $dM^2/M^2$ is not $d\ln M^2$, because the Ito correction from the quadratic variation is dropped. A quick check is $\ln M^2=2\ln M$ with $d\ln M=\sigma\,dW-	frac12\sigma^2dt$, which matches Method 2's $2\sigma\,dW-\sigma^2dt$.

Spotting the faulty step in two Ito derivations

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