Comparing expectations of logarithms of the random variables

Question

Suppose we know that $E(X) > E(Y)$ for $X, Y > 0$. Does the inequality also hold for $E(\log X)$ and $E(\log Y)$?

Of course $\log(E(X)) > \log(E(Y))$. I tried to use the Jensen's inequality but could not really conclude anything.

Answer 1

Let $Y$ is a constant random variable $Y=e$. And $X$ has 2 outcomes: $X=1$ 90% of the time and $X=e^5$ 10% of the time.

$E(X) = 0.9×1+0.1×e^5 > 15$

$E(Y) = e < 3$

$E(\log X) = 0.9×0 + 0.1×5 = 0.5$

$E(\log Y) = 1$

Answer 2

I don't think Jensen's Inequality applies here, but maybe an illustration of why it doesn't apply might help. Before talking about Jensen's Inequality, note that:

• any function $f:\Bbb{R} \to \Bbb{R}$ can be viewed as defining a set $\Bbb{S} \subseteq \Bbb{R}^2$ as $\{(x, f(x)) \mid x \in dom(f)\}$.

• an expectation over a random variable $X$ (or $f(X)$) depends on the choice of probability distribution over its values.

• Let $\Bbb{S}$ be the set of all possible values of $(X, f(X))$, then the convex hull $conv(\Bbb{S})$ are all possible values of $(E[X], E[f(X)])$.

Now, this is the crucial visualization part. Visualize some function, and its convex hull. For any x (drawing a vertical line there), note that f(x) will be lower than every point in the convex hull on that line, if and only if f(x) is convex.

I tried this mental experiment, constructing convex hulls from sets of 2, 3, many, and infinitely many points, for various shapes. And, secondly, convincing myself that any weighted average of the original points cannot escape the convex hull, as long as the weights are a valid probability distribution.

So, back to your question. You have two separate random variables X and Y, and so, in this diagram, the two different underlying probability distributions lead to $E[X] \neq E[Y]$. And, now, note that even if $f(x)$ is increasing, it's still possible for a point to the left in the convex hull of $f(x)$ to be higher than one to the right. They have wiggle room.