# Likelihood Ratio Method - Delta

I was checking Glasserman(2004) - Monte Carlo for Financial Engineering and got to the likelihood ratio method. I am also looking in my textbook (M. Cerrato: The Mathematics of derivatives securities with applications in Matlab). By checking both sources I see some differences and I do not know which one is correct.

Glasserman says that to estimate the BlackScholes delta one has to think of $S(0)$ as a parameter of $S(T)$. We have the density function of a log normally distributed var. $$g(x)=\frac{1}{x\sigma \sqrt{T}}\phi(d(x))$$ where $$d(x)=\frac{ln(x/S(0)-(r-1/2\sigma^2)T}{\sigma \sqrt{T}}$$ so here comes my first question. What is $$\frac{\frac{\partial g(x)}{\partial S(0)}}{g(x)}$$ equal to? I tried it on my own (my mathemtics is not particularly strong so if you are kind to put your calculation step by step I would really appreciate it) and I got $$\frac{\partial\phi(d(x))}{\phi(d(x))}\frac{\partial d(x)}{\partial S(0)}$$ which is quite different from both my lecturer and Glasserman. $$Glasserman=-d(x)\frac{\partial d(x)}{\partial S(0)}$$ $$My\ lecturer=\frac{\partial ln(g(S(T)))}{\partial S(0)}$$ Which of the two is correct (I don't expect mine to be :)) )? and how does the derivation works?

Thank you in advance!

## 1 Answer

The second form is the same as the first where $x=S(T)$. $$\frac{\partial ln(g(x))}{\partial S(0)} = \frac{\partial ln(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial S(0)} = \frac{1}{g(x)}\frac{\partial g(x)}{\partial S(0)}$$ As for the derivation, apply the chain rule. \begin{align} \frac{1}{g(x)}\frac{\partial g(x)}{\partial S(0)} &= \frac{1}{g(x)}\frac{\partial}{\partial S(0)}\left[\frac{1}{x\sigma\sqrt{T}}\phi(d(x))\right] \\ &= \frac{1}{g(x)}\frac{1}{x\sigma\sqrt{T}}\phi'(d(x))\frac{\partial d(x)}{\partial S(0)} \end{align} To continue, note that $\phi'(x) = -x\phi(x)$. I give this without proof and leave it as an exercise for you to verify this.

Now we can finish. \begin{align} \frac{1}{g(x)}\frac{\partial g(x)}{\partial S(0)} &= \frac{1}{g(x)}\frac{1}{x\sigma\sqrt{T}}(-d(x))\phi(d(x))\frac{\partial d(x)}{\partial S(0)} \\ &= -d(x)\frac{\partial d(x)}{\partial S(0)} \end{align}