I have some problem with deciding whether the process $Y_t$ is a Brownian motion. Here is the structure:
$X_t = \int_0^tW_s^2ds$ where $\{W_t\}_{t≥0}$ is a Brownian motion, and $Y_t=\frac{X_t-\frac{1}{2}t^2}{\frac{1}{2}t^{3/2}}$.
So here are what I did:
I calculated the mean and variance for $X$ and $Y$: $\mathbb{E}X_t=\frac{1}{2}t^2$, $\text{Var}X_t= \frac{1}{3}t^4$, $\mathbb{E}Y_t=0$, $\text{Var}Y_t= \frac{4}{3}t$.
I didn’t find the wrong places in my calculations. I found the variance of $Y_t$ is not $t$, so I wanted to say that $Y_t$ is not a Brownian motion. But I feel this argument is not that convincing. I am wondering there might be something wrong but I didn’t notice. Could you give me a convincing proof on whether $Y_t $ is a Brownian motion or not?
Thank you so much!
Best Answer
Note : From Danny's observation, we already know that $Y_t$ cannot be a Brownian motion as it is pathwise differentiable. However, I wish to derive the mean and variance of $Y_t$ for the sake of computation, as OP has demanded this of me.
To calculate the mean and variance of $Y_t$ we need to find the respective quantities for $X_t$.
Now, $E[X_t] = E\left[\int_0^t W_s^2 ds\right]$. We can exchange the integral and expectation since the integrand is non-negative hence Tonelli's theorem applies. Thus, $E[X_t] = \int_0^t E[W_s^2]ds = \int_0^t sds = \frac{t^2}{2}$.
Now, $E[X_t^2] = E\left[\left(\int_0^t W_s^2ds\right)^2\right] = E[\int_0^t W_s^2ds \int_{0}^t W_u^2du] = E\left[\int_0^t\int_0^t W_s^2W_u^2\ du\ ds\right]$. Once again, we have Tonelli applying here, and we get : $$ E[X_t^2] = \int_0^t \int_0^t E[W_s^2W_u^2] duds $$
Now, we will calculate the inner expectation for the case that $s > u$, and then for $u>s$ we can just switch the roles of $u$ and $s$. The advantage of doing that is that we may then write: $$ W_s^2W_u^2 = ((W_s - W_u)+W_u)^2W_u^2 = (W_s-W_u)^2W_u^2 + 2(W_s-W_u)W_u^3 + W_u^4 $$
and therefore by independence and stationarity of increments : $$ E[W_s^2W_u^2] = E[(W_s - W_u)^2]E[W_u^2] + 2E[(W_s - W_u)]E[W_u^3] + E[W_u^4] \\ = (s-u)u + 3u^2 = 2u^2 + us $$
For $u>s$ we have $2s^2+su$. So we have : $$ E[X_t^2] = \int_0^t \left(\int_0^s (2u^2+us)du + \int_s^t (2s^2+us)du\right) ds \\ = \int_0^t \left[\frac{7s^3}{6} + \frac {s}{2}(-5s^2+4st+t^2)\right]ds\\ =\left[\frac 16\left(-2s^4+4s^3t+\frac{3s^2t^2}{2}\right)\right]^t_0 \\ = \frac{7t^4}{12} $$
and therefore $\mbox{Var}[X_t] = E[X_t^2] - E[X_t]^2 = \frac{7t^4}{12} - \frac{t^4}{4} = \frac{1}{3}t^4$, which matches the correct answer.
Now, $E[Y_t] = 2t^{-\frac 32} (E[X_t] -\frac 12 t^2) = 0$.
Then $\mbox{Var}[Y_t] = 4t^{-3}\mbox{Var}(X_t) = \frac {4t}3$, which doesn't match with the variance of a Brownian motion $B_t$ at time $t$.