Expectations of estimators involving ratios

Question

I'm studying some panel data econometric topics, and I've come across the following asymptotic properties as $N \to \infty$: \begin{equation} \theta^{\ast} \to \frac{\mathbb{E}(\beta_i)}{1 - \mathbb{E}(\gamma_i)} \quad \text{and} \quad \bar{\theta} \to \mathbb{E}\left(\frac{\beta_i}{1 - \gamma_i}\right) = \mathbb{E}(\theta_i), \end{equation} The slides state that the two estimators converge to different parameters unless $\beta_i$ and $\gamma_i$ are independently distributed. Why is this the case? Specifically, why is the expected value of a ratio not equal to the ratio of expected values? What role does the independence of the distributions of the random parameters play in this difference? It's simply because if $\beta_i$ and $\gamma_i$ are independent random coefficients, the joint expectation reads: $\mathbb{E} (\beta_i) \times \mathbb{E}\left(\frac{1}{1-\gamma_i}\right)$?

Answer 1

Jensen's Inequality for random variables is involved here.

If, say, $X$ is a random variable and a function $f$ is convex then

$$\mathbb E[f(X)] \geq f\left(\mathbb E[X]\right).$$

The reverse inequality holds if the function is concave.

In the OP's case, suppose that $\beta_i$ and $\gamma_i$ are independent. Then, indeed

$$\mathbb{E}\left(\frac{\beta_i}{1 - \gamma_i}\right) = \mathbb{E} (\beta_i) \cdot \mathbb{E}\left(\frac{1}{1-\gamma_i}\right).$$

Now, consider the function $$f(z) = \frac{1}{1-z}$$

Its second derivative is

$$\frac{\partial ^2 f}{\partial z^2} = 2\frac{1}{(1-z)^3}$$

So whether it is convex or concave (or nothing of the two), depends on the value of $z$. If $z \in (-\infty, 1)$, the second derivative will be positive and the function will be convex. If $z > 1$ the function will be concave. If $z \in (-\infty, \infty)$, then officially the function is neither convex nor concave because the convexity/concavity property is a property defined with respect to the whole domain.

But in general, due to Jensen's inequality,

$$\mathbb{E}\left(\frac{1}{1-\gamma_i}\right) \neq \frac{1}{1-\mathbb E(\gamma_i)}.$$

To provide a way to see this, let's assume that $\gamma_i < 1,\, \forall i$. We can then explore using the sample analogues of the expected value expressions. Let $H(z_i)$ denote the Harmonic mean of $z_i$s, and let $A(z_i)$ denote the Arithmetic mean. Then the sample analogues of the $\mathbb E$-expressions are

$$\mathbb{E}\left(\frac{1}{1-\gamma_i}\right) \approx \frac {1}{H(1-\gamma_i)},\qquad \frac{1}{1-\mathbb E(\gamma_i)} \approx \frac {1}{1-A(\gamma_i)}.$$

Can these be equal?

We inquire whether

$$\frac {1}{H(1-\gamma_i)} =?\; \frac {1}{1-A(\gamma_i)}$$

$$\implies A(\gamma_i) + H(1-\gamma_i) =? \;1. $$

What does hold is that $$A(\gamma_i) + A(1-\gamma_i) = 1.$$

And because, by the Arithmetic-Geometric-Harmonic mean inequality, $$ H(1-\gamma_i) < A(1-\gamma_i)$$

we get that

$$A(\gamma_i) + H(1-\gamma_i) < 1. $$

Answer 2

Specifically, why is the expected value of a ratio not equal to the ratio of expected values?

Suppose $$ (X,Y) = \begin{cases} (1,1) \\ (1,2) \\ (2,1) \\ (2,2) \end{cases} \text{all with equal probabilities: $1/4.$} $$ Then $\dfrac{\operatorname E(X)}{\operatorname E(Y)} = \dfrac{1.5}{1.5} = 1$

and $\operatorname E\left( \dfrac XY \right) = 1.125.$

Since $X,Y$ are independent one can write $\operatorname E\left( \dfrac XY \right) = \operatorname E(X) \operatorname E\left( \dfrac1Y \right).$ Should one then expect that $\operatorname E\left( \dfrac1Y\right) = \dfrac1{\operatorname E(Y)} \text{?}$ One would expect that if the reciprocal function were linear, or if it were affine. But it isn't. Its graph is a curve.

But now consider asymptotics. As the variances of the estimators decrease with increasing sample sizes, one is homing in on a small part of that curve, so it looks like a straight line. Then it may be the case that the expectation of a reciprocal converges to the reicprocal of the expectation. But you still need independence in order to get $\operatorname E\left( \dfrac XY \right) = \operatorname E(X) \operatorname E\left( \dfrac1Y \right).$

For example, suppose $$ (X,Y) = \begin{cases} (1,1) \\ (2,2) \end{cases} \text{ each with probability $1/2$.} $$ Then $\operatorname E\left( \dfrac XY \right) = 1$ and $\operatorname E(X) \operatorname E\left( \dfrac1Y \right) = 1.125.$