# Show that $\mathbb{P}[ a V\le Z| V+Z]=\mathbb{P}[aV \ge Z| V+Z] \text{ a.s.}$ iff $V=\frac{1}{\sqrt{a}}Z'$ where $Z'$ is standard normal

Consider a pair of independent random variables $$(V,Z)$$ where $$Z$$ is standard normal. Now suppose that the following equality holds: for a given $$a>0$$ \begin{align} \mathbb{P}[ a V\le Z| V+Z]=\mathbb{P}[aV \ge Z| V+Z] \text{ a.s.} \end{align}

Question: Can we show that the above is true if and only if $$V=\frac{1}{\sqrt{a}}Z'$$ where $$Z'$$ is standard normal independent of $$Z$$?

To show the forward direction note that if we choose $$V=\frac{1}{\sqrt{a}}Z'$$, we get \begin{align} \mathbb{P} \left[ \sqrt{a} Z'- Z \le 0| \frac{1}{\sqrt{a}}Z'+Z \right]=\mathbb{P} \left[\sqrt{a} Z'- Z \ge 0| \frac{1}{\sqrt{a}}Z'+Z \right] \text{ a.s.} \end{align}

Now note that $$\sqrt{a} Z'- Z$$ and $$\frac{1}{\sqrt{a}}Z'+Z$$ are independent, to see this observe that \begin{align} \mathbb{E} \left[(\sqrt{a} Z'- Z)(\frac{1}{\sqrt{a}}Z'+Z)\right]=0, \end{align} and recall that uncorolated is equivalent to independence for gaussian random variables.

Consequently, we have that \begin{align} \mathbb{P}[ \sqrt{a} Z'- Z \le 0]=\mathbb{P}[\sqrt{a} Z'- Z \ge 0] \end{align} which is true by summary.

The other direction appears to be hard or not true. I am not sure.
I was thinking of re-writing the equality condition as \begin{align} \mathbb{P} \left[ \left[\begin{array}{c} a\\-1 \end{array} \right] (V,Z) \le 0 | \left[\begin{array}{c} 1\\1 \end{array} \right] (V,Z) \right]=\mathbb{P} \left[ \left[\begin{array}{c} a\\-1 \end{array} \right] (V,Z) \ge 0 | \left[\begin{array}{c} 1\\1 \end{array} \right] (V,Z) \right], \end{align} and then either using some kind projection theorem or change of variable.