Expectation of ratio of two identically distributed random variables is greater than $1$: $\mathbb E[X/Y] \ge 1$ for $X \sim Y $

Question

Let $X, Y$ be two identically distributed (i.d.) positive random variables. If they are furthermore independent, from the Cauchy-Schwarz inequality (last step), one has

$$ E[X/Y] = E[X]\cdot E[1/Y] = E[X]\cdot E[1/X] \geq E[\sqrt{X}/\sqrt{X}]^2 = (E[\sqrt{X}/\sqrt{X}])^2 = 1 $$

If one doesn't assume independence of $X$ and $Y$ anymore (but one still assumes that $X\sim Y$), by Jensen inequality one gets $$ \log E[X/Y] \geq E[\log X - \log Y] $$ and the right hand side is $0$ if $\log X$ is assumed integrable, and this gives again $(*)\, E[X/Y] \geq 1$

I am pretty sure (still in the case where $X$ and $Y$ are identically distributed and possibly dependent) that $(*)$ holds true without the hypothesis $\log X \in L^1$, but my attempts don't work (I tried to replace $X$ and $Y$ by truncated versions, but the dominated convergence theorem doesn't work). Maybe there is another way to do it?

Answer 1

Let $X$ and $Y$ be positive, identically distributed RVs. WLOG we may assume $\mathbb{E}[X/Y] < \infty$. Then

Lemma 1. $\mathbb{E}[\log (X/Y)]$ exists in $[-\infty, \infty)$. Moreover, we have $$ \mathbb{E}[X/Y] \geq 1 + \mathbb{E}[\log (X/Y)]. $$

Proof. This is an immediate consequence of the inequality $1 + x \leq e^x$.

Now the key observation is as follows:

Lemma 2. Let $U$ and $V$ be arbitrary positive RVs such that $\mathbb{E}[\log(U/V)]$ exist in $[-\infty, +\infty]$. Then

$$ \mathbb{E}[\log(U/V)] = \int_{0}^{\infty} \frac{\mathbb{E}[e^{-Vs}] - \mathbb{E}[e^{-Us}]}{s} \, \mathrm{d}s. $$

Using Lemma 1 and 2 and noting that $\mathbb{E}[e^{-Ys}] = \mathbb{E}[e^{-Xs}]$, the desired inequality is easily obtained:

$$\begin{align*} \mathbb{E}[X/Y] &\geq 1 + \mathbb{E}[\log(X/Y)] \\ &= 1 + \int_{0}^{\infty} \frac{\mathbb{E}[e^{-Ys}] - \mathbb{E}[e^{-Xs}]}{s} \, \mathrm{d}s \\ &= 1 \end{align*}$$

Remark. When it comes to answering OP's question, Lemma 2 is way too powerful because it applies to a much more general situation. Utilizing the identical distribution condition, we can come up with a more elementary substitute, for example this answer, for our purpose.

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Proof of Lemma 2. For each $x \in \mathbb{R}$, let $x_+ = 0 \vee x$ and $x_- = 0 \vee (-x)$ be the positive and negative part of $x$, respectively. Then by the Frullani's integral and the Tonelli's theorem,

$$\begin{align*} \mathbb{E}[(\log(U/V))_+] &= \mathbb{E}[\log(U \vee V) - \log V] \\ &= \mathbb{E}\left[ \int_{0}^{\infty} \frac{e^{-Vs} - e^{-(U\vee V)s}}{s} \, \mathrm{d}s \right] \\ &= \int_{0}^{\infty} \frac{\mathbb{E}[e^{-Vs}] - \mathbb{E}[e^{-(U\vee V)s}]}{s} \, \mathrm{d}s. \end{align*}$$

By the same reasoning, we also have

$$\begin{align*} \mathbb{E}[(\log(U/V))_-] &= \int_{0}^{\infty} \frac{\mathbb{E}[e^{-Us}] - \mathbb{E}[e^{-(U\vee V)s}]}{s} \, \mathrm{d}s. \end{align*}$$

However, since $\mathbb{E}[\log (U/V)]$ is assumed to exist in $[-\infty, +\infty]$, at least one of the above integrals is finite. Hence, their difference is well-defined, yielding

$$\begin{align*} \mathbb{E}[\log(U/V)] &= \mathbb{E}[(\log(U/V))_+] - \mathbb{E}[(\log(U/V))_-] \\ &= \int_{0}^{\infty} \frac{\mathbb{E}[e^{-Vs}] - \mathbb{E}[e^{-Us}]}{s} \, \mathrm{d}s \end{align*}$$

as required.

Answer 2

By Jensen inequality and the result discussed here, one gets

$$ \log \mathbb E[X/Y] \geq \mathbb E[\log (X/Y)]=\mathbb E[\log X - \log Y]=0 $$

considering that $\log X \sim \log Y$ and $\mathbb E[\log (X/Y)]$ exists (because $\log (X/Y)$ is bounded from above by $X/Y -1$, whose expectation exits by assumption).

Answer 3

One can approximate the integral above with Lebesgue sums. So one could show that every (conveniently chosen) Lebesgue sum has value $\ge 1$. For this, it is enough to check the following: consider a square $n\times n$ matrix $(\rho_{ij})$, $\sum_{i,j} \rho_{ij} = 1$, with the property that for every $1\le i \le n$ we have $\sum_{j} \rho_{i j} = \sum_{j} \rho_{j i}$, and let $x_1$, $\ldots$, $x_n$ be numbers $>0$. Then we have the inequality:

$$\sum_{ij}\rho_{ij} \frac{x_i}{x_j} \ge 1$$

Indeed, the sum is $\ge $ ( mean inequality)

$$\prod_{ij} \left( \frac{x_i}{x_j}\right)^{\rho_{ij}}$$

and this product is in fact $1$ ( check that the exponent of every $x_i$ is in fact $0$)

Note: it may not be the most straighforward proof but it avoids some convergence problems.

Answer 4

Here is a more elegant solution (assuming independent continuous variables with distribution function $f$):

$$ E[\frac{X}{Y}]=\iint_{x<y}\frac{x}{y}f(x)f(y)dx dy+\iint_{y<x}\frac{x}{y}f(x)f(y)dx dy \\\stackrel{symmetry}{=} \iint_{x<y}\frac{x}{y}f(x)f(y)dx dy+\iint_{x<y}\frac{y}{x}f(x)f(y)dx dy\\ =\iint_{x<y}\left(\frac{x}{y}+\frac{y}{x}\right)f(x)f(y)dx dy $$

The AM-GM inequality gives for any $x,y>0$ that $\frac{\frac{x}{y}+\frac{y}{x}}{2}\geq 1$. This yields $$ E[\frac{X}{Y}]\geq \iint_{x<y}2f(x)f(y)dx dy \stackrel{symmetry}{=}\iint_{x,y} f(x)f(y)dx dy =1$$