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In control variate technique we have to calculate $$b=\frac{\text{cov}\{{X,Y}\}}{\text{var}\{{X}\}}$$ where $X$ is a payoff from standard call option and $Y$ is a payoff from for example barrier option. Why we have to estimate $b$ before we use this method and we cant use payoffs which we use during pricing? Or maybe I can first calculate the payoffs for these options, then based on them, calculate the option price, and finally calculate $b$ using the same payoffs and change the price of the barrier option accordingly?

EDIT: My current problem: I try to calculate value of Up and out call option using MC simulation. I use three methods: 1) standard Monte Carlo, 2) anthitetic variates MC, 3) control variates MC using standard call option. The correct price in BS model is $1.3341$. My three methods (with the same parameters and number of simulations equal $100000$) gives me these results:

  1. MC: $1.3621$

  2. Antithetic MC: $1.3763$

  3. Control variates MC: $1.3703$

Is it normal that using standard MC I get the best price?

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Suppose you want to price a derivative $X$ (e.g. a barrier option). You simulate $M$ sample paths and compute $M$ potential discounted payoffs, $f_X$. The standard Monte Carlo estimate for the price of $X$ is simply the (arithmetic) average, $$\text{Price}=\bar{f_X}=\frac{1}{M}\sum_{i=1}^Mf_X(i).$$

The idea of control variates is that you use your $M$ paths to price another derivative, $Y$, which is (very) similar to $X$. For example, if you price a barrier option, then you can use a vanilla option as control variate. Importantly, for this control variate (the derivative $Y$), you need to know a closed-form solution, call it $f_Y^*$.

Now, you have got $M$ sample payoffs for $X$ (denoted by $f_X$) and $M$ sample payoffs for $Y$ (denoted by $f_Y$) as well as the closed-form solution for $Y$ (denoted by $f_Y^*$). Based on your sample payoffs, let's calculate $$\hat\beta = \frac{\mathbb{C}\text{ov}(f_X,f_Y)}{\mathbb{V}\text{ar}[f_Y]}.$$ This looks like a regression coefficient (market beta) or a minimum variance hedge. The new control variate estimate for the price of your derivative is then $$\text{Price}=\bar{f_X}+\hat\beta(f_Y^*-\bar{f_Y}).$$ The term $f_Y^*-\bar{f_Y}$ is the bias of your random numbers. Scaling the bias correction by $\hat\beta$ ensures that the new variance is at smaller (or equal to) the sample variance of $f_X$. The idea is that $\hat\beta$ emerges as optimal (i.e. variance minimising) solution of the problem $$\min_\beta\; \mathbb{V}\text{ar}[f_X+\beta(f_Y^*-f_Y)]=\mathbb{V}\text{ar}[f_X]+\beta^2\mathbb{V}\text{ar}[f_Y]-2\beta\mathbb{C}\text{ov}(f_X,f_Y).$$

Clearly, the more $f_X$ and $f_Y$ correlate, the better the method. Imagine you price a down-and-out put option. This payoff actually correlates negatively with a vanilla put payoff. Not including $\hat\beta$ could make your price estimate worse!

Check out Boyle, Broadie and Glasserman (1997) for more details. For example, you can use several instruments simultaneously as control variates and the optimal $\hat\beta$ coefficients will, of course, look similar to those from a multi linear regression.

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  • $\begingroup$ Thanks for great answer! I edit my question because I dont know why antithetic and control variates doesnt work good for barrier option. $\endgroup$ Commented Jan 24, 2021 at 20:26
  • $\begingroup$ @JohhnWhite Instead of just focussing on the point estimate, you should compute the confidence intervals. They should decrease as you employ more variance reduction methods (hence the name). $\endgroup$
    – Kevin
    Commented Jan 25, 2021 at 9:07

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