weighted sum of two i.i.d. random variables Suppose we know that $X_1$ and $X_2$ are two independently and identically distributed random variables. The distribution of $X_i$ ($i=1,2$) is $P$, and we have some constraints on $P$ that $$\mathbb{E} X_1 = 0$$ (zero-mean) and $$\mathbb{E} X_1^2 = 1$$ (variance is normalized). We denote the set of all feasible distributions as $\mathcal{D}$.
Then, my question is about the function 
$$f(w_1,w_2,a): = \sup_{P\in \mathcal{D}} \Pr(w_1X_1+w_2X_2\ge a)$$
where $w_1\ge 0$, $w_2\ge 0$ and $a> 0$. 
Q1: Can we get an analytic solution of $f?$ 
Q2: Can we know some properties of $f$? For example, it is easy to show that $f$ is monotonically decreasing as $a$ increases when $w_1$ and $w_2$ are fixed. I want to ask if $a$ is fixed, and $w_1+w_2$ is fixed, is $f$ (quasi-)convex (or concave, monotonic...) on $w_1$? 
For the two questions above, I also appreciate answers for another definition of the set $\mathcal{D}:=\{ P \mid \mathbb{E} X_1 = 0, P(|X_1|>1)=0 \text{ (bounded instead of restriction on variance)}\}.$
Thank you in advance.
 A: If you let $a$ and $\omega_1 + \omega_2 =: c$ be fixed and, for notational convenience, let $0 < \omega_1 =: t$, then 
\begin{align}
   f(\omega_1,\omega_2,a)
 = f(t)
&= \sup_{P \in \mathcal{D}}P[t X_1 + (c-t)X_2 \geq a] \\
&= \sup_{P \in \mathcal{D}} \int P\left[X_1 \geq \left(1-\frac{c}{t}\right)X_2 +  
   \frac{a}{t} \mid X_2\right]dP \\
&= \sup_{P \in \mathcal{D}} \left\{ \int_{X_2 < \frac{a}{t-c}} P\left[X_1 \geq 
   \left(1-\frac{c}{t}\right)X_2 + \frac{a}{t} \mid X_2\right]dP \right. \\
&\ \qquad + \left. \int_{X_2 > \frac{a}{t-c}} P\left[X_1 \geq 
   \left(1-\frac{c}{t}\right)X_2 + \frac{a}{t} \mid X_2\right]dP \right\}
\end{align}
For fixed $x_2$, the term $\left(1-\frac{c}{t}\right)x_2 + \frac{a}{t}$ converges to $0$ as $t$ increases. It does so monotonically, from below if $x_2 > \frac{a}{t-c}$ and from above otherwise, and consequently the set 
$\{X_1 \geq (1-\frac{c}{t})x_2 + \frac{a}{t} \}$ decreases or increases respectively.  Therefore, the first integrand will almost surely increase with $t$, and the second will almost surely decrease. 
From this and the moment conditions alone, it doesn't seem possible to conclude monotonicity of $f$. However, if $|X_2| < 1$ holds a.s., then if $a \geq \omega_2$ only the first integrand contributes and one can say that $f$ is increasing with $t$. If $a \leq - \omega_2$, the reverse conclusion can be drawn.
It isn't clear that the increase/decrease should be monotonic though, so I'll hedge and say non-decreasing/non-increasing.
