Question: Assume we have the following equation:

$$\widehat{\Theta}(\rho) = \frac{1}{(1-\rho)\Delta t} \ln\left(\frac{1}{T} \sum_{t=1}^T \left(\frac{1+r_t}{1+rf_t}\right)^{1-\rho} \right) \ \ \ \ \cdots \ (1)$$


$T =$ total number of observations

$\Delta t =$ length of time between observations

$r_t =$ a variable with random values at each $t$

$rf_t =$ a variable with random values at each $t$

$\rho =$ a constant 'coefficient'

(Note: for those familiar with finance terminology, $r_t$ is an asset's return at time $t$ and $rf_t$ is the risk-free rate at time $t$.)

Computing $\widehat{\Theta}$ does not require any specific distribution for $r_t$ or $rf_t$. I have a large sample of data (i.e, $T = 162,401$) for $r_t$ and $rf_t$.

Now define $\hat{\rho}$ where $\hat{\rho}$ satisfies $\widehat{\Theta}(\hat{\rho}) = 0$. Solving for $\hat{\rho}$ is relatively straightforward using Newton–Raphson method or other linear optimization methods. My aim is to derive the asymptotic distribution for $\hat{\rho}$, that is, as $T \rightarrow \infty$:

  1. What can we say about the asymptotic distribution of $\hat{\rho}$?

  2. Can we derive a particular distribution to which it converges to?

  3. Can we derive it's asymptotic variance/mean?

  4. What other asymptotic properties can we derive?

Attempt: My attempt so far has been related to a 2nd order linear approximation to Eqn.$(1)$ rather than working directly with Eqn.$(1)$. Using Taylor series expansion and properties of the Generalized Mean, we can show that:

$$\widehat{\Theta}(\rho) \approx \frac{1}{\Delta t} \left[\overline{x} + \frac{1-\rho}{2} \left(s_x^*\right)^2\right] \ \ \ \ \ \cdots \ (2)$$


$\overline{x} = \frac{1}{T} \sum_{t=1}^T x_t$ where $x_t = r_t - rf_t$

$\left(s_x^*\right)^2 = \left(\frac{T-1}{T}\right)s_x^2$ where $s_x^2 = \frac{1}{T-1}\sum_{t=1}^T (x_t - \overline{x})^2$

From Eqn.$(2)$, an approximation to $\hat{\rho}$ is given by:

$$\hat{\rho} \approx \frac{2\overline{x}}{\left(s_x^*\right)^2} +1 $$

Since there is no closed form solution to $\hat{\rho}$ from Eqn.$(1)$, how can we derive 'approximate' asymptotic properties of $\hat{\rho}$ using Eqn.$(2)$?

  • 1
    $\begingroup$ Have you tried doing MC simulations to see the how distribution looks like? $\endgroup$
    – mpiktas
    Nov 19, 2013 at 12:01
  • $\begingroup$ @mpiktas, I haven't actually, but simulating $\widehat{\rho}$ should be pretty straightforward. However, I was more interested if there is some kind of rigorous proof (or references to similar cases) to derive the asymptotic distribution mathematically. $\endgroup$
    – TeTs
    Nov 19, 2013 at 12:27
  • $\begingroup$ If the resulting distribution is normal, you can then just think how to apply CLT creatively, which is relatively easy, since there are a lot of similar results to learn from. If the result is not normal then the problem is more complicated. $\endgroup$
    – mpiktas
    Nov 20, 2013 at 7:48

1 Answer 1


$$\hat{\rho} \approx \frac{2\overline{x}}{\left(s_x^*\right)^2} +1$$ $$\Rightarrow \frac 12 (\hat{\rho} -1) \approx \frac{\overline{x}}{\left(s_x^*\right)^2} = [\left(s_x^*\right)^2]^{-1}\overline{x} = \Big(\frac{1}{T}\sum_{t=1}^T (x_t - \overline{x})^2\Big)^{-1}\cdot \Big(\frac{1}{T} \sum_{t=1}^T x_t\Big)$$

If we can assume that the process $\{X_t\}$ is weakly stationary, with mean $\mu_x$ and variance $\sigma^2_x$ then, by applying the relevant CLT we can quickly arrive at

$$\hat{\rho} \sim_{approx} N\left(1+ \frac {2\mu_x}{\sigma^2_x}, \frac 4{\sigma^2_xT}\right) $$

for "large samples".

To clarify some questions that emerged in the comments:
Assuming conditions for CLT holds, then define the variable $W=\bar x -\mu_x$ and obtain

$$\sqrt TW \rightarrow_d N(0, \sigma^2_x)$$

and then as an approximation, we have

$$W \sim_{approx} N(0, \sigma^2_x/T)$$

for large samples. Now substitute in the equation for $\hat \rho$ the expression $\bar x = W+ \mu_x$ to obtain

$$\hat{\rho} \approx \frac{2}{\left(s_x^*\right)^2}(W+ \mu_x) +1 = \frac{2}{\left(s_x^*\right)^2}W+ \Big(\frac{2}{\left(s_x^*\right)^2}\mu_x +1\Big)$$

which, given the approximate distributional result on $W$, gives you the approximate distribution for $\hat \rho$, for large samples, using of course Slutsky as pointed out by @mpiktas. And yes, these results are full of approximations.

  • $\begingroup$ Thanks, how exactly did you arrive at the final distribution? I know that $\overline{x} \stackrel{p}{\rightarrow} \mu_x$ and $\left(s_x^*\right)^2 \stackrel{asymp}{\sim} N(\sigma_x^2, \frac{\mu_4-(\sigma_x^2)^2}{n}) $ where $\mu_4 = E[(x_t-\mu_x)^4]$. $\endgroup$
    – TeTs
    Nov 20, 2013 at 5:16
  • $\begingroup$ Alecos used the fact that $\sqrt{T}\bar{x}\to N(0,\sigma_x^2)$ and $s_x^2\xrightarrow{P}\sigma^2$ in probability. Then you can employ Slutsky's lemma, if $X_n\to X$ in distribution and $Y_n\to a$ in probability with $a$ being the constant, then $X_nY_n\to aX$ in distribution. $\endgroup$
    – mpiktas
    Nov 20, 2013 at 7:51
  • $\begingroup$ @mpiktas, Actually I just want to clarify, why does $\sqrt{T}\overline{x} \rightarrow N(0, \sigma_x^2)$? I thought by the CLT, we have $\frac{\sqrt{T}(\overline{x} - \mu_x)}{\sigma_x} \rightarrow N(0,1)$, so $\sqrt{T}\overline{x} \rightarrow N(0, \sigma_x^2)$ only if we assume $\mu_x = 0$? Why can we make that assumption? Thanks. $\endgroup$
    – TeTs
    Nov 21, 2013 at 10:07
  • $\begingroup$ I wrote for clarity, not for correctness. Yes in general you need to subtract the mean to get assymptotic normality. $\endgroup$
    – mpiktas
    Nov 21, 2013 at 11:05

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