Inverse GBM Moments and Simulation

Article Quant Q&A · Author: Cindy Philip

Summary

The document considers the reciprocal of a stock price modeled by geometric Brownian motion (GBM). It gives the standard GBM solution and uses the moment-generating identity for a Gaussian variable to derive the reciprocal’s mean. For initial price S₀, drift μ, volatility σ, and time t, that mean is S₀⁻¹ exp((-μ + σ²)t), rather than the proposed S₀⁻¹ exp(μt). The reciprocal’s variance follows from its second moment: S₀⁻² exp((-2μ + 2σ²)t)(exp(σ²t) − 1).

The simulation question also exposes a coding error: updating the stock and separately adding a GBM-like increment to its inverse does not simulate the reciprocal process. Instead, simulate the stock path and take its reciprocal at the endpoint, or simulate the transformed process using its correct dynamics. The document gives formulas and a simulation example, but does not provide a corrected implementation or a numerical validation of the reciprocal moments.

Key ideas

The reciprocal of a GBM price is lognormal, with its log drift adjusted by the volatility term.
The expected reciprocal is S₀⁻¹ exp((-μ + σ²)t), not S₀⁻¹ exp(μt).
The reciprocal variance can be derived by subtracting the squared mean from its second moment.
An inverse price should be computed from the simulated stock value or modeled with the reciprocal’s correct dynamics.

Tags

Statistics Volatility Backtesting

Full text

# Simulation of Geometric Brownian Motion


# Simulation of Geometric Brownian Motion












I generate 10000 random binomial paths for a stock whose price is from S(0) = 10 out to S(t) where t = 1 year. Assume geometric Brownian motion for the stock price with a drift of 15% per year and a volatility of 20%. I use 10000 equally spaced time steps of length along each path and $\Delta S = \mu S \Delta t \pm \sigma S \sqrt{\Delta t}$, where the + or – moves at each time step are generated at random with equal probability.

```
S = 10
S1 = []
mean = 0.15
sd = 0.2

for i in range(10000):
    S = 10

    for j in range(10000):
        roll = np.random.rand()
    
        if roll < 0.5:
            S = S + mean*S/10000 + sd*S/100
        else:
            S = S + mean*S/10000 - sd*S/100

    S1.append(S)
```

I use the above codes to simulate the GBM and my result is the same as the theoretical values. Mean should be $S(0)e^{\mu t}$ and the variance should be $S^2(0)e^{2\mu t}(e^{\sigma^2 t} - 1)$. But there is one question want me to calculate the inverse, which is $G(t) = 1/S(t)$. I consider that mean should be $\frac{1}{S(0)}e^{\mu t}$ and variance should be $\frac{1}{S^2(0)}e^{2\mu t}(e^{\sigma^2 t} - 1)$. Am I correct? But I'm not sure how to change my codes. I have tried to write down the following codes. But the result is far from the theoretical values mean = 0.116 and variance = 0.00055. Could someone explain where I'm doing wrong? Thank you very much.

```
S = 10
G1 = [1/10]
mean = 0.15
sd = 0.2

for i in range(10000):
    S = 10

    for j in range(10000):
        roll = np.random.rand()
    
        if roll < 0.5:
            G = 1/S
            S = S + mean*S/10000 + sd*S/100
            G = G + mean*(1/S)/10000 + sd*(1/S)/100
        else:
            G = 1/S
            S = S + mean*S/10000 - sd*S/100
            G = G + mean*(1/S)/10000 - sd*(1/S)/100

    G1.append(G)
```

## Answer by Magic is in the chain (score 3, accepted)

https://quant.stackexchange.com/a/61571

Just to explain the formulae for the mean and variance! We can start with the solution of the GBM SDE:

$S_t=S_0\,e^{\left(\mu-\frac{1}{2}\sigma^2\right)t+\sigma W_t}$

then,

$\frac{1}{S_t}=\frac{1}{S_0}\,e^{-\left(\mu-\frac{1}{2}\sigma^2\right)t-\sigma W_t}$

The mean (and variance) are easily obtained via this identity for a Gaussian random variable Y:

$E\left[e^Y\right]=e^{E\left[Y\right]+\frac{1}{2}V\left[Y\right]}$

Applying this to S, and noting that:

$E\left[\left(\mu-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right]=\left(\mu-\frac{1}{2}\sigma^2\right)t$

$V\left[\left(\mu-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right]=\sigma^2t$

we get the familiar expression for the mean of S:

$E\left[S_t\right]=S_0\,e^{\left(\mu-\frac{1}{2}\sigma^2\right)t+\frac{1}{2}\sigma^2 t}=S_0\,e^{\mu t}$

And applying the identity to the expression for 1/S, we get its mean as follows:

$E\left[\frac{1}{S_t}\right]=\frac{1}{S_0}\,e^{-\left(\mu-\frac{1}{2}\sigma^2\right)t+\frac{1}{2}\sigma^2 t}=\frac{1}{S_0}\,e^{-\mu t+\sigma^2t}$

Variance can be determined using the same identity.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.

Inverse Geometric Brownian Motion: Moments and Simulation

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