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I recently came across a question whether a Monte Carlo simulation should represent a forward curve at each tenor. I encountered an approach at a bank which I would consider as somehow strange.

Approach 1: Let’s take GBPEUR as an underlying. To keep things as simple as possible I would choose a Geometric Brownian Motion with constant drift and diffusion calibrated to forward volatilities on GBPEUR. In the next step I would choose an Euler Scheme, simulate 100000 paths in order to solve the stochastic differential equation numerically. By taking the mean over the paths at each tenor I get a point which is close to the forward curve at this tenor. But in my opinion it’s nearly impossible to calibrate the Geometric Brownian Motion in a way that it exactly represents the forward GBPEUR curve on each tenor. Nevertheless that’s the way a simulation is done!

So what I now saw implemented in a financial institution is the following.

Approach 2: The guy took each tenor of the forward curve and multiplied it with a normally distributed random number. I would say this should be represented like this:

            F_t=F_t N(0,σ) 
            F_t GBPEUR forward curve at tenor t
            N(0,σ) a normally distributed random number with zero mean  and σ the volatility 

Then he took 100000 draws and said that’s my Monte Carlo simulation and of course trivially his simulation represented in the mean the whole forward curve.

So here are my questions:

  • Do you agree that the approach 1 is the right one? Of course there are technically more advanced ones, but this is the way it is done.
  • Do you agree that it is nearly impossible to represent the whole forward curve with a Monte-Carlo simulation regardless of how complex the underlying sdes are? Otherwise we would have a perfectly calibrated model.
  • Approach 2 is awfully wrong and has nothing to do with a Monte Carlo Simulation.

Thanks in advance

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You are confusing Monte Carlo which is a numerical method and the modelling that comes before that.

The 2 approaches use Monte Carlo simulation but it is the modelling that is different.

In approach 2, the forwards for different expiries are modelled as completely independent. The corresponding process is a white noise. This is not realistic at all but this is not an issue in itself if what you are interested in does not depend on the covariance between rates. For example, you could get the right value for a portfolio of vanilla options at the expiries you simulate. Obviously you would't need a MC simulation for that and a much more serious issue is that your FX forward can actually be negative but I hope you get my point.

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  • $\begingroup$ Thanks AFK, I think I get the point. So the questions is whether my proposed model for the underlying is realistic in the sense that it captures the behaviour of the underlying in the best way possible.So I am going to run a lot of statistics on the time series for my underlying in regards to trend, autocorrelation etc. ... I then use the MC Simulation as a tool regardless whether the model is right or wrong in order to get e.g. a price $\endgroup$ – BirthdayPony Oct 19 '16 at 20:48
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Assume deterministic interest rates to keep notations uncluttered.

In the absence of arbitrage opportunity, under the risk-neutral measure $\Bbb{Q}$ associated to a risk-free money market account denominated in EUR, you suggest to model the GBP/EUR spot exchange rate dynamics using the following SDE: $$ \frac{dX_t}{X_t} = (r^{EUR}_t - r^{GBP}_t)dt + \sigma dW_t^{\Bbb{Q}},\quad X(0)=X_0$$ Solving the equation yields the unique solution: \begin{align} X_T &= \underbrace{X_0 \exp\left( \int_0^T (r^{EUR}_t - r^{GBP}_t) dt \right)}_{=X(0,T)} \underbrace{\exp(- \frac{1}{2}\sigma^2 T) \exp(\sigma W_T^\Bbb{Q})}_{= \mathcal{E}[W_t]} \tag{0} \end{align} with $X(0,T)$ denoting the forward exchange rate and $\mathcal{E}[X_t] = \exp(X_t - \frac{1}{2}\langle X \rangle_t)$ the stochastic exponential of a continuous semi martingale $X_t$. From the properties of the latter exponential, it is straightforward to see that: $$ X(0,T) = \Bbb{E}^\Bbb{Q} \left[ X_T \vert \mathcal{F}_0 \right] \tag{1} $$

When you are evaluating a forward price using Monte Carlo simulations you are numerically evaluating $(1)$ as: $$\tilde{X}(0,T) = \frac{1}{M} \sum_{m=1}^M X_T^{(m)} \tag{2} $$ where $\{ X_T^{(m)} \}_{m=1}^M$ are iid samples of the random variable $X_T$ simulated under the risk-neutral measure $\Bbb{Q}$, conditionally on the knowledge of $X_0$.

Assuming you manage to sample from the true distribution of $X_T$ (which is possible under our working modelling assumptions), then central limit theorem (CLT) tells you that your Monte Carlo estimation of the forward exchange rate, although unbiased, will have a variance $\propto M^{-1/2}$. This is probably what you mean by "you can never exactly represent the forward price". But I don't understand the approach 2 you mention: do you mean a plus sign instead of a multiplication?

That being said, you do not need Monte Carlo to compute forward prices: in general, those are computable in closed form as I just showed. Also, yes, you can perfectly calibrate a GBM to match a given forward cruve as long as your model can accomodate a term structure for the risk-neutral drift. It is basically equivalent to matching the first moment of the conditional distributions.

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  • $\begingroup$ Hi Quantuple, thank you very much for your elaborate answer. I totally agree with you. Unfortunately approach two has now plus sign in its equation. I saw it like this implemented in a risk management system at a fund. I think that the most fundamental problem with approach 2 is, that it does not represent a SDE. Thus it is just numerical noise around each tenor of the forward curve. $\endgroup$ – BirthdayPony Oct 19 '16 at 14:38
  • $\begingroup$ @BirthdayPoiny Hi. Sorry but what do you mean exactly by "around each tenor of the forward curve"? If there is no plus sign how can you have the right expectation, $F(0,t)$ is not a random variable it is a number, so $F(0,t) N(0,\sigma^2)$ is a RV with an expectation of zero. $\endgroup$ – Quantuple Oct 19 '16 at 14:45
  • $\begingroup$ fair enough, I see your point. So let's assume that we have F(0,t)=F(0,t)+N(0,sigma^2) $\endgroup$ – BirthdayPony Oct 19 '16 at 15:03
  • $\begingroup$ In my opinion in order to get a proper simulation we must at least assume something like F(0,t+s)=F(0,t)+N(0,sigma^2) for one timestep from t to s. Otherwise we are going to get nothing more than numerical noise... $\endgroup$ – BirthdayPony Oct 19 '16 at 15:08
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    $\begingroup$ @John, I simply meant that the forward curve describes a term structure of first moments, that of the risk neutral distribution of the underlying $q(t,s)=d\Bbb{Q}(S_t\leq s)/ds$. Nothing more. In that respect, matching a forward curve = matching the first moment of future prices' distributions (under RN mesure $\Bbb{Q}$). $\endgroup$ – Quantuple Oct 19 '16 at 16:56

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