$W_t$ is a brownian motion and $X_t:= \sqrt{t}Z$, where: $Z\sim N(0,1)$.

How to show that for a bounded continuous $f$ process, $$U_t = \int_0^t (f(W_s))ds$$ and $$V_t = \int_0^t (f(X_s))ds$$ have the same expectation but not the same variance in general?

  • $\begingroup$ I am not sure to get the point of your question. Basically, Brownian motion and your $X$ stochastic process are equal in distribution. So all their moments will be equal, as well as as the moments of your two processes $U$ and $V$. They can be ‘different’ in the sense of non-indistinguishability (a.s. equality), just as two identically distributed random variables can be different ‘event-wise’ $\endgroup$
    – siou0107
    Commented Apr 20, 2020 at 15:07
  • 3
    $\begingroup$ The thing though is that the $X_t$ process is very smooth, while $W_t$ is very rough. The quadratic variation will be very different. $\endgroup$
    – nbbo2
    Commented Apr 20, 2020 at 15:55
  • $\begingroup$ I don’t agree with that. $dX_t = \sqrt{dt}Z$, which has the same distribution and properties (such as quadratic variation) than $dW_t$ $\endgroup$
    – siou0107
    Commented Apr 20, 2020 at 17:15

1 Answer 1


The means are equal

Suppose $f$ is analytic so that we can give it a Taylor series that works everywhere such that $f(x) = \sum a_n x^n$, and then let us let this be bounded too. To show that the expectations are the same consider taking the expectation of $U_t$ \begin{equation} \mathbb{E}(U_t) = \mathbb{E}\left(\int_0^t f(W_s) \,\mathrm{d}s\right)\\ \end{equation} as $f$ if bounded we can use Fubini's theorem to move the expectation inside the integral, giving \begin{align} \mathbb{E}(U_t) & = \int_0^t \mathbb{E}\left(f(W_s)\right) \,\mathrm{d}s\\ & = \int_0^t \sum_n a_n \mathbb{E}\left(W_s^n\right) \,\mathrm{d}s \\ & = \int_0^t \sum_n a_n s^{n/2} M_n \,\mathrm{d}s \end{align} where $M_n$ denotes the $n$-th moment of a standard Gaussian distribution. Identically we could do this for the expectation of $V_t$ and see that the two are equal, and hence that $U_t$ has the same expectation of $V_t$.

The variances are not equal

For the variance of $U_t$ we need to evaluate the expected value of \begin{align} U_t^2 = \left(\int_0^t f(W_s) \,\mathrm{d}s \right)^2 &= \int_0^t \int_0^t f(W_s) f(W_u) \,\mathrm{d}u \,\mathrm{d}s \\ &= 2 \int_0^t \int_0^s f(W_s) f(W_u) \,\mathrm{d}u \,\mathrm{d}s. \end{align}

Taking the expectation and putting this inside the integral again gives \begin{equation} \mathbb{E}\left(U_t^2\right) = 2 \int_0^t \int_0^s \sum_{n, m, k} a_n a_m \binom{n}{k} \mathbb{E}(W_s^n W_u^m) \,\mathrm{d}u \,\mathrm{d}s. \end{equation} Within the integrand $u \leq s$ and so expand $W_s = W_u + (W_s - W_u)$ where as $W$ is a Weiner process the increment $W_s - W_u$ is independent to the value of $W_u$ and has variance $s - u$. This then simplifies to give \begin{equation} \mathbb{E}\left(U_t^2\right) = 2 \int_0^t \int_0^s \sum_{n, m, k} a_n a_m \binom{n}{k} u^{\frac{n + m}{2}}(s - u)^{\frac{n-k}{2}}M_{n+m}M_{n-k} \,\mathrm{d}u \,\mathrm{d}s. \end{equation}

An identical process can be applied for $V_t$ but instead of expanding $W_s$ we just use $X_t = \sqrt{t}Z$ and obtain \begin{equation} \mathbb{E}\left(V_t^2\right) = 2 \int_0^t \int_0^s \sum_{n, m, k} a_n a_m \binom{n}{k} s^{\frac{n}{2}}u^{\frac{m}{2}}M_{n+m} \,\mathrm{d}u \,\mathrm{d}s. \end{equation}

The second expression is more easily integrated out than the first, but both are doable and I think they evaluate to different quantities.

A numeric demonstration

Take $f \equiv \sin$ and you can simulate this, as the example python code does, the key result is then:

Variable = U,   Mean = 0.00206326180593 +/- 0.00228420660865,   Standard deviation = 0.228420660865 
Variable = V,   Mean = -0.0177668651186 +/- 0.0380135632954,    Standard deviation = 3.80135632954 

We see this confirms that the means are the same but variances differ.

enter image description here

Python code

import numpy as np
from scipy.stats import norm

f = np.sin

M = 10000
T = 2*np.pi
dt = T/M
N = 10000
mean_u, mean_v = 0, 0
var_u, var_v = 0, 0
for _ in range(N):
    w = np.cumsum(np.concatenate([[0], norm.rvs(scale=dt, size=M)]))
    z = norm.rvs()
    t = np.cumsum([0] + [dt]*M)
    x = np.sqrt(t) * z
    u = np.sum(f(w))*dt
    v = np.sum(f(x))*dt
    mean_u += u
    mean_v += v
    var_u += u ** 2
    var_v += v ** 2
mean_u /= N
mean_v /= N
var_u /= N
var_v /= N
var_u -= mean_u ** 2
var_v -= mean_v ** 2
std_u = np.sqrt(var_u)
std_v = np.sqrt(var_v)
std_error_u = np.sqrt(var_u / N)
std_error_v = np.sqrt(var_v / N)

for n, m, s, se in [["U", mean_u, std_u, std_error_u], ["V", mean_v, std_v, std_error_v]]:
    print("Variable = {},\tMean = {} +/- {},\tStandard deviation = {} ".format(n, m, se, s))
  • $\begingroup$ Your $X$ process does not seem much random to me... It is just a random variable which, once known, grows at $\sqrt{t}$, but then it is not a stochastic process. $\endgroup$
    – siou0107
    Commented Apr 24, 2020 at 15:15
  • $\begingroup$ That's exactly the definition of the process $X_t$ in the original post: $X_t:=\sqrt{t}\times Z$. $\endgroup$ Commented Apr 24, 2020 at 15:17
  • $\begingroup$ @siou0107 It's no more or less random than $W_t$. Once a sample $\omega$ is realised both $X_t$ and $W_t$ are known, it's just that one is much smoother than the other. $\endgroup$
    – oliversm
    Commented Apr 24, 2020 at 15:48
  • $\begingroup$ @oliversm Thanks so much for your answer... $\endgroup$
    – Alicia
    Commented Apr 27, 2020 at 0:15
  • 1
    $\begingroup$ The quadratic variation of the process $X_t := \sqrt{t} Z$ will be zero. Take the definition of the quadratic variation it follows $\langle X\rangle_t := \lim_{\Delta t \to 0} \sum_k (X_{t_k} - X_{t_{k-1}})^2 = \lim \sum (\sqrt{t_k}Z - \sqrt{t_{k-1}}Z)^2 = Z^2 \lim \sum (\sqrt{t_k} - \sqrt{t_{k-1}})^2 = 0$, where the last equality follows from the quadratic variation of the sqrt root function (which is a finite variation function). $\endgroup$
    – oliversm
    Commented Jan 26 at 19:05

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