5

I just read a proof in a text that first established that

$$(X-\mu_X)(Y-\mu_Y)\le \frac12\left[(X-\mu_X)^2+(Y-\mu_Y)^2\right]$$

then took the expectation of both sides of the inequality.

$$E[(X-\mu_X)(Y-\mu_Y)]\le \frac12E\left[(X-\mu_X)^2+(Y-\mu_Y)^2\right]$$

How is taking the expectation of both sides of the inequality justified?

David
  • 2,262
  • 2
    Depending on your technical definition of probability, this is either obvious or tedious, but the root of the matter is that the if the variable on the left is always less than the variable on the right, then so is the expected value. – Callus - Reinstate Monica Jun 18 '14 at 23:58
  • 5
    Expectation is linear, so you can subtract one side from both to simplify the statement you need to the basic $X\geq 0\implies E[X]\geq 0$. From here there are any number of approaches, but they all boil down to the fact that you're 'adding' (or integrating, which amounts to the same thing) a number of non-negative quantities, and the results of such addition/integration will always be non-negative. – Steven Stadnicki Jun 18 '14 at 23:59
  • 4
    It is true that if $X\lt Y$ then $E(X)\lt E(Y)$. Maybe it will be more obvious in the form if $Y-X\gt 0$ then $E(Y-X)\gt 0$. – André Nicolas Jun 19 '14 at 00:01
  • Nice replies everyone, very, very helpful. – David Jun 19 '14 at 00:58

1 Answers1

6

It is known that $\mathbb{E}[X]\leq \mathbb{E}[Y]$. This comes from the definition of expected value on a probability space $(S, \mathcal{G}, P)$ as

$$ \mathbb{E}[X] = \int_{S}XdP, $$

where the integral is the Lebesgue integral. This property can be verified using the definition of the Lebesgue integral.

Now in your question you have also asked whether $X<Y$ (almost surely) implies $\mathbb{E}[X]<\mathbb{E}[Y]$ (with $<$ instead of $\leq$). This is also true: Suppose that $X<Y$ and $\mathbb{E}[X]=\mathbb{E}[Y]$. Define a random variable $T=Y-X$; this is nonnegative and

$$ 0\leq \mathbb{E}[T]=0. $$

As a result, $T=0$ a.e., therefore, $X=Y$ a.e., which is a contradiction (we have used the fact that for any nonnegative random variable $T$, $\mathbb{E}[T]=0$ implies that $T=0$ a.s.).

user322903
  • 716