$E(X_t^2)=tE(X_1^2)$ for a martingale $X$ with stationary and independent increments The result is clear if $t$ is an integer larger than $1$, since then for all $s<t$,  $E(X_t^2\mid\mathcal F_s)=E((X_t-X_s)^2+2X_tX_s-X_s^2\mid\mathcal F_s) = E(X_{t-s})^2+X_s^2$ (with last equality by stationarity of increments and $X$ a martingale), so $E(X_t^2)=E(X_{t-s})^2+E(X_s^2)$. In particular, let $s=1$ and apply the equality repeatedly.
I'm not sure how to get the result for $t\ge0$. Any suggestions are greatly appreciated.
Thanks in advance.
 A: You have more than the result for integer $t$. 
Consider $E[X_t^2] = E[X_{t-s}]^2 + E[X_s^2]$ with $t=\alpha s$. You find $E[X_{\alpha s}^2] = E[X_{(\alpha-1)s}]^2 + E[X_{s}^2]$ for any $\alpha \geq 1$ and $s \geq 0$. In particular, for $\alpha = n \in \mathbb{N}$,
$$
E[X_{ns}^2] = E[X_{(n-1)s}^2] + E[X_s^2]
$$
which by induction implies
$$
E[X_{ns}^2] = nE[X_{s}^2].
$$
Most notably, this applies not just for integer $s$, but all $s \geq 0$.
In particular, take $t=n/m\cdot s$ with $n,m$ integers.
Then the previous formula implies
$$
nE[X_t^2] = E[X_{nt}^2] = E[X_{m\cdot (\frac{n}{m}\cdot t)}^2] = m E[X_{\frac{n}{m}\cdot t}^2].
$$
Rearranging gives
$$
E[X_{\frac{n}{m} t}^2] = \frac{n}{m}E[X_t^2].
$$
Finally, you will need some kind of continuity (right or left continuity is fine). Then given any $\alpha\geq 0$ approximate it by rationals $r_n$ so that
$$
E[X_{\alpha t}^2] = \lim_n r_n E[X_t^2] = \alpha E[X_t^2].
$$
Careful how to justify pulling the limit out here (use that $X^2$ is a submartingale). Now take $t=1$.
