# Upper bound on Brownian motion increments

My question is regarding Brownian motion and its upper bound.

Brownian motion can be simulated by partition the interval $$[0,T]$$ into $$N$$ subintervals with step size $$\Delta t = \frac{T}{N}$$ such that $$t_0 = 0, t_1 = \Delta t, \ldots, t_i = t_{i-1}+ \Delta t, \ldots, t_N = T.$$ Then the increments of Brownian motion can be simulated as $$B(t_i)-B(t_{i-1}) = \sqrt{t_i-t_{i-1}}Z_i, ~~~~i=1,\ldots,N$$ where $$Z_i$$ are independent random variables with distribution $$\mathcal{N}(0, 1)$$. By definition we have $$B(t_i)-B(t_{i-1})\sim \mathcal{N}(0, t_i-t_{i-1})$$ which equals $$\sqrt{t_i- t_{ i-1}} \mathcal{N}(0, 1)$$. The variable $$Z_i$$ is a standard normal distribution, i.e. it reaches its maximum value at $$\frac{1}{\sqrt{2\pi}}$$.

Consider a stochastic differential equation $$dX_t = X_t dt + \alpha B_t,$$ where $$\alpha>0$$ and $$B_t$$ is Brownian motion. The Euler-Maruyama simulation gives $$X_{i+1} = X_{i} + X_{i}\Delta t + \alpha \cdot (B_{t_{i+1}} - B_{t_i}).$$

The diffusion term of the simulated equation $$X_{i+1}$$ is then bounded in relation to the maximum value of the normal distribution. But Brownian motion is of infinite variation. How can this be when we can prove that the increments are bounded?

• I didn't know that the normal distribution has a maximum value. Its pdf is non zero on the entire $\mathbb R$. Mar 27, 2022 at 9:48

## 1 Answer

The probability density of the normal distribution has this maximum at zero, not the random variable that is normally distributed. That random variable has a positive probability density at any value, meaning that it is unbounded.

• Yes that's right, I was confused for a moment. Thanks for your help! Mar 28, 2022 at 8:42