Lemma 1.2.2.label Let $(\Omega, \cf, \bp)$ be a probability space, $E$ be a separable Banach space over $\real$, and $X, X': \Omega \to E$ be i.i.d Gaussian random variables, then $((X + X')/\sqrt{2}, (X - X')/\sqrt{2})$ have the same distribution as $(X, X')$.
Proof. Let $x^{*}, y^{*} \in E^{*}$, $Y(\omega) = \angles{X(\omega), x^*}_{E}$ and $Y'(\omega) = \anglesn{X'(\omega), x^*}_{E}$, then $Y$ and $Y'$ are independent Gaussian random variables, so
\begin{align*}\ev^{\bp}\braks{e^{i\angles{X, x^*}_E + i\anglesn{X', y^*}_E}}&= \ev^{\bp}\braks{e^{i\angles{X, x^*}_E}}\ev^{\bp}\braks{e^{i\anglesn{X', y^*}_E}}\\&= \exp\braks{-\frac{1}{2}\paren{\ev^\bp\braks{\angles{X, x^*}_E^2} + \ev^\bp\braks{\angles{X', y^*}_E^2}}}\end{align*}
On the other hand,
\begin{align*}&\ev^{\bp}\braks{e^{i\anglesn{(X + X')/\sqrt{2}, x^*}_E + i\anglesn{(X - X')/\sqrt{2}, y^*}_E}}\\&= \ev^{\bp}\braks{e^{i\anglesn{X/\sqrt{2}, x^* + y^*}_E}}\ev^{\bp}\braks{e^{i\anglesn{X'/\sqrt{2}, x^* - y^*}_E}}\\&= \exp\braks{-\frac{1}{4}\paren{\ev^\bp\braks{\angles{X, x^* + y^*}_E^2} + \ev^\bp\braks{\angles{X', x^* - y^*}_E^2}}}\end{align*}
where
\begin{align*}\angles{X, x^* + y^*}_{E}^{2}&= \angles{X, x^*}_{E}^{2} + \angles{X, y^*}_{E}^{2} + 2\angles{X, x^*}_{E}\angles{X, y^*}_{E} \\ \angles{X', x^* - y^*}_{E}^{2}&= \angles{X', x^*}_{E}^{2} + \angles{X', y^*}_{E}^{2} - 2\angles{X', x^*}_{E}\angles{X', y^*}_{E}\end{align*}
Since $X$ and $X'$ are identically distributed,
\begin{align*}&\ev^{\bp}\braks{\angles{X, x^* + y^*}_E^2}+ \ev^{\bp}\braks{\angles{X', x^* - y^*}_E^2}\\&= \ev^{\bp}\braks{\angles{X, x^*}_E^2 + \angles{X', x^*}_E^2}+ \ev^{\bp}\braks{\angles{X, y^*}_E^2 + \angles{X', y^*}_E^2}\\&+2\ev^{\bp}\braks{\angles{X, x^*}_E\angles{X, y^*}_E}- \ev^{\bp}\braks{2\angles{X', x^*}_E\angles{X', y^*}_E}\\&= 2\ev^{\bp}\braks{\angles{X, x^*}_E^2}+ 2\ev^{\bp}\braks{\angles{X', y^*}_E^2}\end{align*}
Thus the characteristic functions of $(X, X')$ and $((X + X')/\sqrt{2}, (X - X')/\sqrt{2})$ are equal. By Definition 1.1.5, their distributions are equal.$\square$