Suppose $V$ and $W$ are independent. Let \begin{align} f(x)=E[ ((1-x)V+xW )^p] \end{align} for $x \in [0,1]$ and some even $p \ge 2$. Find \begin{align} \min_{0 \le x \le 1} f(x) \end{align}
Also, suppose that $E[W^{2n+1}]=0$ for $2n+1 <p$, that is all odd moments of $W$ are zero. (I think this should simplify this a little bit)
Solution, to some of the cases can be done by expanding and taking the derivative.
Case of $p=2$ \begin{align} E[ ((1-x)V+xW )^2]= (1-x)^2 E[V^2]+x^2 E[W^2] \end{align} it is not difficult to check that the minimizing $x= \frac{E[V^2]}{E[V^2]+E[W^2]}$ and the minimum value is given by \begin{align} \min_{x\in [0,1]} f(x) =\frac{E[V^2]E[W^2]}{E[V^2]+E[W^2]} \end{align}
Case of $p=4$
\begin{align} E[ ((1-x)V+xW )^2]= (1-x)^4 E[V^4]+6x^2(1-x)^2E[V^2]E[W^2]+x^4 E[W^4] \end{align}
and one must solve \begin{align} -4(1-x)^3 E[V^4]+6E[V^2]E[W^2] (2x-5x+x^3)+4x^3E[W^4]=0 \end{align} which I guess can be done.
But this procedure becomes increasingly complicated.
Are there any other techniques on how to approach this problem? I also feel that this should have come up in some applications?
Here is also an example when the problem can be solve in general
The problem can be solved if $V \sim \mathcal{N}(0,\sigma_v^2)$ and $V \sim \mathcal{N}(0,\sigma_w^2)$. Since, \begin{align} E[ ((1-x)V+xW )^p] =E \left[( \sqrt{(1-x)^2\sigma^2_V+x^2\sigma^2_W} Z )^p \right] = ((1-x)^2\sigma^2_V+x^2\sigma^2_W)^{p/2} E[Z^p] \end{align} where $Z$ is standard norma. It is not difficult to check that the expression is minimized by $x=\frac{\sigma_V^2}{\sigma_V^2+\sigma_W^2}$ (same as for $p=2$)
and we get that \begin{align} \min_{x \in [0,1]} f(x)= \frac{(\sigma_V^2\sigma_W^2)^{p/2}}{(\sigma_V^2+\sigma_W^2)^{p/2}} E[Z^p] \end{align}
Upper Bound
We can derive the following upper bound on the problem by using value inspired optimal value for $p=2$ and triangle (Minkowski) inequality. We use $x=\frac{E^{2/p}[V^p]}{(E^{2/p}[V^p]+E^{2/p}[W^p])}$ \begin{align} \min_{x\in [0,1]} E[ ((1-x)*V+x W )^p ] &\le \frac{ E \left[ (E^{2/p}[W^p] V +E^{2/p}[V^p] W )^p \right])}{ (E^{2/p}[W^p]+E^{2/p}[V^p])^p} \\ &\le \frac{ \left(E^{2/p}[W^p] E^{1/p}[V^p] +E^{2/p}[V^p] E^{1/p}[W^p] \right)^p }{ (E^{2/p}[W^p]+E^{2/p}[V^p])^p}\\ &= \frac{E[W^p]E[V^p] ( E^{1/p}[W^p]+E^{1/p}[W^p])^p}{(E^{2/p}[W^p]+E^{2/p} [V^p])^p}\\ & \le 2^p \frac{E[W^p]E[V^p]}{(E^{2/p}[W^p]+E^{2/p} [V^p])^{p/2}} \end{align}