Likelihood Ratio Estimator for Black–Scholes Delta
Summary
The document explains a step in the likelihood ratio method for estimating Black–Scholes delta by Monte Carlo. Treating the initial asset price as a parameter of the terminal-price density, the method uses the derivative of the log density with respect to that parameter as a weighting term. The answer clarifies that differentiating the log density is equivalent to dividing the density derivative by the density itself.
For a lognormal terminal-price density, the prefactor does not depend on the initial price, so the derivative passes through the normal density’s argument. Using the identity that the derivative of the standard normal density is the negative of its argument times the density yields the likelihood ratio score, expressed as minus the standardized terminal-price variable times its derivative with respect to the initial price. The explanation relies on the stated density parameterization and chain rule; it does not derive the full Monte Carlo estimator or discuss variance reduction and implementation details.
Key ideas
- The likelihood ratio method differentiates the terminal-price density with respect to the initial price.
- The log-density derivative equals the density derivative divided by the density.
- The chain rule applies through the normal density’s standardized argument.
- The standard normal density derivative supplies a negative argument factor.
- The result is a score term that can weight simulated payoffs when estimating delta.
Tags
Full text
# Likelihood Ratio Method - Delta
# 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!
## Answer by msitt (score 1)
https://quant.stackexchange.com/a/33979
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. $$ \begin{align} \frac{1}{g(x)}\frac{\partial g(x)}{\partial S(0)} &= \frac{1}{g(x)}\frac{1}{x\sigma\sqrt{T}}(-d(x))\phi(d(x))\frac{\partial d(x)}{\partial S(0)} \\ &= -d(x)\frac{\partial d(x)}{\partial S(0)} \end{align} $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.