I have attempted the following question but fear I may have sort of perhaps made my solution fit the answer for convenience sake. I have the following:
Y is Brownian motion with variance parameter $\sigma^2$. Prove the process X with $$ X(t) = Y\left(\frac{t}{\sigma^2}\right) $$ is a standard Brownian motion
My attempt is as follows:
We know: $ \mbox{Var}(Y(t)) = E[\{Y(t+s) - Y(s)\}^2] = \sigma^2t$
This is a known result. We now try to prove $\mbox{Var}(X(t))=t$ $$ \begin{eqnarray*} \mbox{Var}(X(t)) & = & E[\{X(t+s) - X(s)\}^2] \\ & = & E\left[\left\{Y(\frac{t+s}{\sigma^2}) - Y(\frac{s}{\sigma^2})\right\}^2\right] \\ &=& E\left[ \left\{ \frac{1}{\sigma}Y(t+s) - \frac{1}{\sigma}Y(s) \right\}^2 \right] \\ &=& E\left[\frac{1}{\sigma^2} \left\{ Y(t+s) - Y(s) \right\}^2 \right]\\ &=& \frac{1}{\sigma^2} E\left[ \left\{ Y(t+s) - Y(s) \right\}^2 \right] \\ &=& \frac{1}{\sigma^2}( \sigma^2 t)\\ &=& t \end{eqnarray*} $$ Thus we prove it has variance of a standard Brownian process. This is the only difficult part I struggled in proving and think I may have perfomed steps that aren't true such as the third step. The rest of the requirement for standard Brownian motion were simple to prove. Please may I ask for your guidance and assistance in understanding how to prove this. I also messed around with the idea of fidgeting with the transition function but...did not progress far.
Your th ird step can be justified as follows: it is enough to show that $Y(\frac {t+s} {\sigma^{2}})-Y(\frac {s} {\sigma^{2}})$ has the same distribution as $\frac 1 {\sigma} Y( t+s)-\frac 1 {\sigma} Y(s)$. We know that $Y(\frac {t+s} {\sigma^{2}})-Y(\frac {s} {\sigma^{2}})$ has the same distribution as $Y(\frac {t} {\sigma^{2}})$ and $\frac 1 {\sigma} Y( t+s)-\frac 1 {\sigma} Y(s)$ has the same distribution as $\frac {Y(t)} {\sigma}$. But $Y(t)$ is normal so $Y(\frac {t} {\sigma^{2}})$ has the same distribution as $\frac {Y(t)} {\sigma}$.