Expected Squared Sum of Two Wiener Process Values

Summary

The document evaluates the expected square of the sum of two Wiener process increments from a common starting time. It expands the square into two individual second moments and a cross-product, then uses the covariance structure of Brownian motion: the variance of an increment over an interval is its length, and the covariance of values at times s and t is the earlier time. This gives s + t + 2 min(s,t), assuming the starting value W₀ is zero or that the process is handled through increments.

A second answer challenges the question’s stated solution and mean for a stochastic differential equation, noting that the exponential moment of Brownian motion must be accounted for. The document does not develop that correction, and the displayed SDE solution appears to contain transcription or notation errors, so it is best treated as a brief covariance calculation rather than a complete treatment of the SDE.

Key ideas

• Expand the squared sum into two variances and a cross-product term.
• For standard Brownian motion, the covariance of Wₛ and Wₜ is the earlier of the two times.
• The resulting expectation for the increments is s + t + 2 min(s,t).
• The accompanying SDE solution and mean are challenged but not fully corrected in the discussion.

Tags

# How to compute $\mathbb{E} \left[ (W_s + W_t - 2W_0)^2 \right]$?


# How to compute $\mathbb{E} \left[ (W_s + W_t - 2W_0)^2 \right]$?












The solution to the SDE

$$dx_t= -kx_t dt + cx_t dW_t$$

is

$$x_t = x_0 e^{\left(c - \frac{k^2}{2} \right)t}e^{-k W_t}$$

with mean

$$\mathbb{E} \left[ x_t \right] = x_0 e^{\left(c - \frac{k^2}{2}\right)t}$$

where $W_)$ is the Wiener process.

I'm looking to compute

$\mathbb{E} \left[ (W_s + W_t - 2W_0)^2 \right]$

but am unsure of how to proceed.

## Answer by aajajim (score 8)

https://quant.stackexchange.com/a/9757

I would calculate it this way,

$\mathbb{E}[(W_s+W_t−2W_0)^2] = \mathbb{E}\left[\left((W_s-W_0)+(W_t-W_0)\right)^2\right]\\ \hspace{4cm}=\mathbb{E}[(W_s-W_0)^2]+\mathbb{E}[(W_t-W_0)^2]+2\mathbb{E}[(W_s-W_0)(W_t-W_0)] \\ \hspace{4cm}=s+t+2\mathbb{E}[W_sW_t]\\ \hspace{4cm}=s+t+2\min(s,t)$

## Answer by wsw (score -1)

https://quant.stackexchange.com/a/9762

Your solution $x_t$ is wrong.

Your mean is wrong too. Note that $\mathbb{E}\left[ e^{W_t}\right] = e^\frac{t}{2}$. I corrected the typo that was pointed out by Richard.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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