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Hi, In my research, I have the following problem. Let $X,Y$ two i.i.d random variables and a function $u(x)=x^2$ if $x>0$ and $ u(x)= -\beta\ (-x)^2$ if $x\leq{}0$ with $\beta\geq{}1$

I need to find the conditions such that there is

$\alpha\in{}(0,1)$

that maximizes

$Eu(\alpha X+(1-\alpha) Y)$.

Thanks

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    $\begingroup$ The convoluted way you have written the definition for $u(x)$ when $x\le0$ makes me suspect a typing mistake. Also, what do you mean by “the conditions”? I suspect it can be tricky to find necessary and sufficient conditions, so a stronger hint of what exactly you need could be helpful. $\endgroup$ May 9, 2010 at 21:40
  • $\begingroup$ By conditions, do you mean conditions on the distribution of X (and therefore Y)? $\endgroup$ May 10, 2010 at 0:06
  • $\begingroup$ Yes, I think there must be some conditions on the cumulative distribution of X and Y. $\endgroup$
    – megozcue
    May 10, 2010 at 0:45
  • $\begingroup$ One way to study this problem is to think under what conditions of the distributions of X and Y, $\alpha=0$ or $\alpha=1$ can not be an optimum. $\endgroup$
    – megozcue
    May 10, 2010 at 1:07
  • $\begingroup$ What happen with the maximum if the function is changed to: $ u(x)=(x)^{0.5}$ if $x>0$ $ u(x)= -\beta\ (-x)^{0.5}$ if $x\leq{}0$ with $\beta\geq{}1$. $\endgroup$
    – megozcue
    May 28, 2010 at 21:59

1 Answer 1

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This is just a comment, not an answer. (I don't have enough reputation to write comments.)

Some basic statements that you probably knew already:

If $P(X \leq 0) = 1$, then there will always be such $\alpha$. Likewise if $P(X \geq 0) = 1$, there will never be such $\alpha$. This is from concavity/convexity of $u(x)$.

If the law of X is symmetric about zero, there will always be such $\alpha$. In particular $\alpha = 1/2$ will be better than $\alpha = 0$ or 1.

I agree with Harald that a general answer may be too much to hope for. Are you more interested in sufficient conditions or necessary conditions? Is there a particular family of distributions that you care about? Do you expect $\beta$ to be very close to 1, or much larger?

If you're looking for a sufficient condition, maybe you could let

$f(\alpha) = E[u(\alpha X + (1-\alpha) Y)]$

and look for situations in which $f'(0) > 0$.

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