Given $Y(t)=\frac{1}{7}B(49t),\:t\ge0$. $B$ denotes brownian motion position.
Have to show that $E(Y(t)^{2k+1})=0,\:k\in\mathbb{N} $.
So far, I've done showing that $Y(t)$ is a Brownian motion. Then I've tried using the formula for mathematical expectation $E(X)=\int_{-\infty}^{\infty}xf(x)dx$.
Letting $x\equiv B(49t)$, $$E(Y(t))=\int\limits_{-\infty}^{\infty} \frac{1}{7}x\frac{1}{\sqrt t\sqrt{2\pi}}\exp{-\frac{1}{2}(\frac{x/7}{\sqrt t})^2} dx $$
What to do next to advance?
Am I on the right track?
Use induction and integration by parts. Suppose that it is true for k = m. Then (I'm omitting constants that are irrelevant for the final result below):
$F(m) = \int\limits_{-\infty}^{\infty}x^{2m+1} e^{-x^2} = \frac{1}{2m+1}\left( x^{2m+2}e^{-x^2}|_{-\infty}^{\infty} + \int_{-\infty}^{\infty}x^{2m+3}e^{-x^2}\right)$
$ = 0 + \frac{1}{2m+1}F(m+1)$
Thus if the result is true for k = m, it is also true for k = m+1. We know that it is true for k = 0, so the result follows.