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Minimisation over random variables

  1. Apr 10, 2014 #1
    Suppose we have a function ##F:\mathbb{R}_+\to\mathbb{R}_+## such that ##\frac{F(y)}{y}## is decreasing.
    Let ##x## and ##y## be some ##\mathbb{R}_+##-valued random variables.
    Would ##\mathbb{E}x\leq\mathbb{E}y## imply that ##\mathbb{E}F(x)\leq\mathbb{E}F(y)##?
     
  2. jcsd
  3. Apr 10, 2014 #2

    jbunniii

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    Suppose we take ##F(y) = 1/y##. Then certainly ##F(y)/y = 1/y^2## is decreasing for positive ##y##.

    Now let ##x## be some random variable which is restricted to the interval ##[1,2]## and let ##y## be some other random variable restricted to the interval ##[3,4]##. Thus ##E[x] < E[y]##. But ##F(x)## is restricted to ##[1/2, 1]## and ##F(y)## is restricted to ##[1/4, 1/3]##. So ##E[F(y)] < E[F(x)]##.
     
  4. Apr 10, 2014 #3
    Your example makes perfect sense.

    But what if we assume that ##F(x)## is increasing in ##x## and ##F(0)=0##?
     
  5. Apr 10, 2014 #4

    jbunniii

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    What if we take
    $$F(x) = \begin{cases}
    \sqrt{x} & \text{ if }0 \leq x \leq 1 \\
    1 & \text{ if } x > 1
    \end{cases}$$
    Then ##F(x)/x## is decreasing for all positive ##x##.

    Let ##x## be uniformly distributed over ##[1,2]##. Let ##y## be 0 or 3, each with probability 1/2. Then ##E[x] = E[y] = 1.5##.

    But ##F(x) = 1## with probability 1, so ##E[F(x)] = 1##. And ##F(y)## is 0 or 1, each with probability 1/2, so ##E[F(y)] = 1/2##.

    If you want ##F## to be strictly increasing, then give it a tiny positive slope for ##x > 1##, and define ##x## and ##y## as above. The result will still be ##E[F(y)] < E[F(x)]##.
     
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