Using the Distribution Function Technique and then linking it to the normal distribution

Question

I have $Z$ as a standard random variable (with zero mean and unit variance), for all $t≥0$ and $X_t=\sqrt t Z$

I need to find the distribution function ${F_X}_t = P(X≤ x)$ and then show that $X_t $ ~ $ N(0:t)$ for all $t ≥0$ which in essence is me finding that $X_t$ is a normally distributed variable with variance t and zero mean.

So far I have done: ${F_X}_t (x) = P(X_t ≤x)=P(\sqrt t Z ≤x)=P(t≤ (\frac{x}{z})^2$

From here would I then integrate $(\frac{x}{z})^2$ between the bounds of t ?

Am I on the right lines? Any help is greatly appreciated.

Answer 1

In general if $U$ is a random variable and $V=cU$ where $c$ is a positive constant then $$F_V(x)=P(V\leq x)=P(cU\leq x)=P(U\leq x/c)=F_U(x/c)$$