Consider a random sample from a distribution with the p.d.f. (for some $c > 0$)

\begin{equation} f(x) = cx^2 (1 − x), \text{if } 0 < x < 1; \end{equation}

\begin{equation} 0, \text{ elsewhere}. \end{equation}

What is an approximate distribution of $\bar{X}_n$ ? For a sample of size 45, find approximate mean and variance for $\bar{X}_n$ ? Find an approximate distribution of $\bar{X}^3_n$.

How should I go about solving this problem ? More importantly, I'd like to have an intuition of this question. I'm new to statistics so more details would help me more. This may be naive/dumb but we know that we are taking a sample from uniform distribution and why would we want to find its distribution again and how is it leading to an another distribution ? I'm very confused.


First of all, the p.d.f $f$ corresponds to a beta distribution with parameters $\alpha=3$ and $\beta=2$ (hence, not a uniform distribution). Now, by the central limit theorem, $\frac{{\sum\nolimits_{i = 1}^n {X_i } - n \mu }}{{\sigma \sqrt n }}$ converges in distribution to the ${\rm N}(0,1)$ distribution, where $\mu = {\rm E}(X_1)$ and $\sigma^2 = {\rm Var}(X_1)$. This means that, for $n$ sufficiently large (for example, $n=45$), we may write $ \sum\nolimits_{i = 1}^n {X_i } \approx n \mu + \sigma \sqrt n Z$, where $Z \sim {\rm N}(0,1)$. Dividing by $n$ thus gives $\bar X_n \approx \mu + (\sigma/{\sqrt{n}})Z$. Thus, $\bar X_n$ is approximated by the normal distribution with mean $\mu$ and small variance $\sigma^2/n$ (this is exactly the distribution of the right-hand side, $\mu + (\sigma/{\sqrt{n}})Z$). As for the mean and variance of $\bar X_n$, this is a simple exercise, and you should be able to show that the mean is $\mu$ and the variance is $\sigma^2/n$. Finally, to find approximate distribution of $\bar X_n^3$, you may use the approximation $\bar X_n \approx \mu + (\sigma/{\sqrt{n}})Z$.

  • $\begingroup$ It was enough to find the mean and the variance using the formula's in beta distribution. I tried to find $\bar{X}_n^3$ your way but I didn't quite succeed but it was lot easier to use the delta method to solve the problem and finally substituted the mean and variance that I got from the second part if we want to find the exact values. It finally came to around $\bar{X}_n^3$ ~ N(0,0.04). $\endgroup$ – 0x0 Feb 9 '11 at 20:42
  • $\begingroup$ I'm not very sure of myself in delta method. A friend of mine gave me an hint to use delta method for this solution and thus the answer I derived. $\endgroup$ – 0x0 Feb 9 '11 at 20:50

Shai Covo's answer shows you how to find the solutions to your homework problems; the key idea is to use the central limit theorem. My answer is going to address your big picture question about intuition behind why one would care about the central limit theorem and about the distribution of the sample mean.

The central limit theorem says that the mean of a sufficiently large random sample from any distribution with finite variance has approximately a normal distribution. This is an extremely surprising and powerful result. It is probably the most important theoretical result in statistics.

It's surprising because many simple transformations of random variables don't have "nice" (i.e., named) distributions. Even adding two independent and identically distributed random variables from the same distribution rarely yields a nice distribution, and when it does, it yields something different depending on the original distribution (e.g., the sum of two geometrics is negative binomial, the sum of two exponentials is gamma, the sum of two bernoullis is binomial). But the central limit theorem says that if you add up enough independent random variables from the same distribution, regardless of what that distribution is, the resulting sum has approximately a normal distribution. (Again, the original distribution must have finite variance.) There is no a priori reason that one would expect a result like this.

The central limit theorem is powerful because knowing the approximate distribution of the sample mean makes it much easier to use the sample to draw inferences about the original population. To take an example, suppose the water quality in a stream is good if its mean oxygen content is about 5 mg per liter. Suppose your random sample of 45 observations has a mean of 4.62 mg per liter. Assuming no flaws in your sampling design, the difference between what your sample produces and what you were hoping it would produce is either due to chance or due to the stream actually having poorer water quality. Knowing that the sample mean has approximately a particular normal distribution allows you to say what the probability is that the difference is due to chance. If, say, $\sigma = 0.92$ mg per liter, then you can determine that the probability that the difference is due to chance is about 0.28%. Thus it's extremely likely that the difference is not due to chance and that the stream's water quality actually is lower than it should be. (What I've described here is what's called a test of significance. There are other inference procedures that the sample mean can be used for, too. You should know that in practice, the picture with statistical inference is generally more complicated than what I've described here. For instance, you normally don't know $\sigma$, which causes more problems. But this is the basic idea.)

  • $\begingroup$ Actually, you both (Mike & Shai) gave correct answers to my questions perfectly but I had to choose only one. Considering the time and relavence I chose Shai's. I appreciate your help. $\endgroup$ – 0x0 Feb 7 '11 at 16:43
  • $\begingroup$ @Sunil: You're welcome. And if I had to choose one of these two answers to accept, I would have chosen Shai's as well. It more directly answers your question. $\endgroup$ – Mike Spivey Feb 7 '11 at 19:08

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