Question

Why is $E([\Sigma_{i=0}^{n-1}((\Delta_i^nW)^2-{T\over n})]^2) = \Sigma_{i=0}^{n-1}E[((\Delta_i^nW)^2-{T\over n})^2]$ in this solution?

score 42 · Answer 1 · 2018-12-16 04:30:51

Why is $E([\Sigma_{i=0}^{n-1}((\Delta_i^nW)^2-{T\over n})]^2) = \Sigma_{i=0}^{n-1}E[((\Delta_i^nW)^2-{T\over n})^2]$ in this solution?

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List Question

Why is $E([\Sigma_{i=0}^{n-1}((\Delta_i^nW)^2-{T\over n})]^2) = \Sigma_{i=0}^{n-1}E[((\Delta_i^nW)^2-{T\over n})^2]$ in this solution?

What is a white noise ? What is the derivate of the Brownian motion?

Find constants $a$ and $b$ such that $X(t)$ is a Brownian Motion

PDE with terminal condition, Feynman-Kac

Diffusion equation for Brownian motion with a systematic drift component

If $ S_t $ follows a log-normal Brownian motion, what SDE does the square of $ S_t $ follow?

Expectation value of the average displacement squared in a random walk

Inferring observation time from a Brownian motion

Brownian motion, stochastic process

Expected value of Brownian motion at a time decided by a rate one Poisson process.

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Compute $E(M_σ)$ when $σ = \inf(t ≥ 0: |B_t| = 1)$ and $M_t = 4B^2_t +e^{4B_t−8t}−4t$

Compute $E\left((B_t−1)^2\int ^t_0(B_s+1)^2 dB_s\right)$, where $(B_t)$ is a standard Brownian motion

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