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Suppose $X_1,\ldots,X_n$ are i.i.d. $\mathcal U(0,1)$. I am looking for the asymptotic distribution of $$T_n = \prod_{i=1}^n [e{X_i}]^{1/\sqrt{n}} \>.$$

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    $\begingroup$ What attempts have you made at solving this problem? (You might consider taking the approach I did in my first comment: study the logarithm of the product, which is explicitly a sum of random variables. What is its expectation? What is its variance? What theorem(s) do you know about the asymptotic distribution of a sum of iid variables?) $\endgroup$ – whuber Dec 17 '12 at 22:14
  • $\begingroup$ @whuber ♦ I think I got the idea from your comment. Thanks! $\endgroup$ – qomo Dec 17 '12 at 22:25
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    $\begingroup$ If you obtain a satisfactory answer yourself, we would be grateful if you would share it here: it's ok to answer your own question. $\endgroup$ – whuber Dec 17 '12 at 22:34
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Since the OP never answered his (or her) own question:

Following whuber's sensible advice, we intend to consider the logarithm of the thing. First we note that since $X_i \sim U(0,1)$ we have that

$$Z = -\ln X_i \Rightarrow f_z(z) = e^{-z}$$ i.e. $Z$ follows an Exponential distribution with $E(Z) = 1, \text{Var}(Z) = 1$. So the sample mean $\bar Z_n$ from an i.i.d. sample has $E(\bar Z_n) = 1, \text{Var}(\bar Z_n) = 1/n$. Keeping these in mind we plunge into

$$\ln T_n = \ln\left[\prod_{i=1}^n [e{X_i}]^{1/\sqrt{n}} \right] = \ln\left[(e^{1/\sqrt{n}})^n \prod_{i=1}^n {X_i}^{1/\sqrt{n}} \right]$$

$$=\sqrt n\ln e + \frac1{\sqrt n}\sum_{i=1}^n\ln X_i = \sqrt n\left(1-\frac 1n\sum_{i=1}^n\left(-\ln X_i\right)\right) $$

$$\Rightarrow \ln T_n = \sqrt n\left(1-\frac 1n\sum_{i=1}^nZ_i\right) = -\frac{\bar Z_n-E(\bar Z_n)}{SD(\bar Z_n)} $$

The final expression is the subject matter of the classical Central Limit Theorem (the minus sign does not bother us). Since $X$'s are i.i.d. so are their functions, and then

$$\ln T_n \xrightarrow{d} T^* \sim N(0,1)$$

Using the Continuous Mapping (Mann-Wald) theorem $Y_n \xrightarrow{d} Y \Rightarrow h(Y_n) \xrightarrow{d} h(Y)$ for $h$ continuous and setting $h(y) = e^y$ we obtain

$$e^{\ln T_n} = T_n \xrightarrow{d} e^{T^*}$$

Since $T^*$ is a standard normal random variable, $e^{T^*}$ is a log-normal RV, so

$$T_n \xrightarrow{d} LN(0,1),\;\; E(T^*) = e^{1/2}, \text{Var}(T^*) = e^2 -e $$

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