Consider the 2 following approaches to pricing a security:

Monte-carlo ($\mathbb{Q}$-measure)

$\begin{equation} C = \frac{1}{N} \sum_{i=1}^{n} e^{-rT} max(S_i(t) - K, 0) \end{equation}$

Monte-carlo ($\mathbb{P}$-measure)

$\begin{equation} C = \frac{1}{N} \sum_{i=1}^{n} D_i max(S_{i}(t) - K, 0) \end{equation}$

where $D_i = \left(\frac{q_i}{p_i}\right) \frac{1}{(1+r)^t}$ is the stochastic discount factor / deflator / pricing kernel under scenario $i$.

Furthermore, I assume that

\begin{align} ln(S_t/S_0) &\sim N(r - 0.5\sigma^2, \sigma\sqrt{T}) \mbox{ under the $\mathbb{Q}$ measure} \\ ln(S_t / S_0) &\sim N(\alpha - 0.5\sigma^2, \sigma\sqrt{T}) \mbox{ under the $\mathbb{P}$ measure} \end{align} I am wondering how to calculate $q_i$ and $p_i$. Then I would like to compare $C$ under the Black-Scholes, Monte-carlo $\mathbb{Q}$ measure and Monte-carlo $\mathbb{P}$ measure.

I expect that $C_{BS}(S=100, \ K=100, \ \sigma=0.25, \ r=0.03,\ T=1) = 11.35$ is close to $C_{MC, \mathbb{Q}} \ (N=10,000)$ and $C_{MC, \mathbb{P}} \ (N=10,000)$.

I use the standard Monte-Carlo technique - i.e. I simulate values of $S(1)$ by taking a random draw from a normal distribution and converting this to the appropriate lognormal distribution.

  • $\begingroup$ This post has helped me understand pricing under P measure. Thank you. Just one question: how/where do you get Di=(qi/pi) * 1/(1+r)^t $\endgroup$
    – Jay Na
    Commented Apr 23, 2017 at 4:12

2 Answers 2


Essentially the question is importance sampling $$ \int f(S_T) \psi_{r}(S_T) dS_T = \int f(S_T) \psi_{\alpha}(S_T) \frac{\psi_r}{\psi_\alpha}(S_T) dS_T $$ Here $\psi_{\mu}$ denotes the log-normal density with drift $\mu.$ So when you simulate with drift $\alpha$ each sample used is $$ f(S_T) \frac{\psi_r}{\psi_\alpha}(S_T) $$ instead of $f(S_T).$

You can also do a change of variable and write the integral in terms of $\log S_T$ to get simpler expressions for the densities.



I think I figured it out.

Under the $\mathbb{Q}$ measure,

$\begin{equation} S_t \sim LN(ln(S_0) + r - 0.5\sigma^2, \ \sigma\sqrt{t}) \end{equation}$

Under the $\mathbb{P}$ measure

$\begin{equation} S_t \sim LN(ln(S_0) + \alpha - 0.5\sigma^2, \ \sigma\sqrt{t}) \end{equation}$

Suppose we simulate the following stock prices

Si = 96.33, 69.04, 115.19

The corresponding probabilities of observing these prices under the $\mathbb{Q}$ measure are

qi = 0.0164, 0.00777, 0.0118

The corresponding probabilities of observing these prices under the $\mathbb{P}$ measure are

pi = 0.01615, 0.006879, 0.01228

The stochastic discount factors are $\left( \frac{q_i}{p_i} \right) e^{-r}$

 Di = 0.9848683 1.0956499 0.9301017

We can then compute the Monte-carlo price of the option under the $\mathbb{P}$ measure using

$\begin{equation} \dfrac{1}{3} \sum_{i=1}^{3} max(0, S_i - K) D_i \end{equation}$

  • $\begingroup$ 2 questions: how do you determine $\alpha$, and how do you define probability of observing $S_i$ under $\mathbb{Q}$ for a continuous distribution? $\endgroup$
    – crunch
    Commented Jul 14, 2015 at 11:11
  • $\begingroup$ $\alpha$ can be estimated from the historical stock prices. Suppose we've observed $S_1, \dots, S_n$ and calculated $x_1, \dots, x_{n-1}$ where $x_i = ln(S_{i+1} / S_{i})$. Let $\mu$ represent the mean of the $x_i$ terms and $s^2$ represent the variance of the $x_i$ terms. Then $\alpha$ can be estimated by $\mu + 0.5 s^2$. For question 2, since we made the assumption that $S_t \sim LN( )$ under the $\mathbb{Q}$ measure, we have the probability density function, $P(S = S_t)$. $\endgroup$
    – nathanesau
    Commented Jul 14, 2015 at 15:08
  • $\begingroup$ But for any continuous distribution, the probability of observing any specific value is 0. $\endgroup$
    – crunch
    Commented Jul 15, 2015 at 7:01
  • $\begingroup$ There's no need to be condescending. For $X$ continuous, $Pr[a \le X \le b] = \int_a^b f(x) dx$. If $a=b$ then that probability is 0. $\endgroup$
    – crunch
    Commented Jul 15, 2015 at 9:50
  • $\begingroup$ I'm sorry my last comment was little rude. I'll try to explain why the probability of observing any specific value is technically zero, but that isn't what a pdf describes exactly. Consider a continuous cumulative distribution function $F$. $F(-\infty)$ = 0 and $F(\infty) = 1$. Define $f(x) = \frac{d}{dx} F(x)$. Clearly $f(x)$ is not equal to 0 at all points. In fact $f(x)$ can be interpreted as the probability that $x$ lies within the area near $x$ $\endgroup$
    – nathanesau
    Commented Jul 15, 2015 at 17:32

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