Is $E[W^2_{t}|F_s]$ a Martingale?

Question

G'day, I am trying to find out mathematically if $W^2_{s}$ is a martingale.

Let $W^2_{s}$ be a Brownian motion, t>_0

$E[W^2_{t}|F_s] = E[(W_t-W_s +W_s)^2|F_s]$

$= E[(W_t-W_s)^2 - W^2_{s} +2(W_t-W_s)W_s|F_s]$

$*E[(W_t-W_s)^2] = Var[(W_t-Ws)] = t-s$

$= E[(t-s) - W^2_{s} +2(W_t-W_s)W_s|F_s]$

$* E[W^2_{s}] = s$

$= E[(t-s) - s +2(W_t-W_s)W_s|F_s]$

$* 2(W_t-W_s)W_s = 0?

$= E[(t-s) - s + 0|F_s]$

Can someone finish/correct what I started?

Answer 1

If $(W_t)$ is supposed to be Brownian Motion then $(E(W_t^{2}|\mathcal F_s))_{t \geq s}$ is not a martingale. If it is, then $E(E(W_t^{2}|\mathcal F_s))$ would not depend on $t$ but in this case, it is $t$.

[If $(X_t)$ is a martingale then $EX_{t+s}=EX_t$ for all $t,s$ so $EX_t$ is independent of $t$].