Suppose $X_{1}, X_{2}, ....$ is a random sample from a probability distribution with mean $\mu < \infty$ and $\sigma^{2} < \infty$. Let $\bar{X}_{n} = \frac{\sum X_{i}}{n}$. From the Central Limit Theorem we can state that: $\sqrt{n}(\bar{X}_{n} - \mu) \xrightarrow {d} N(0,\sigma^{2})$

All right, but now, i want to know how to explain (intuitively) the next statements:

$(\bar{X}_{n} - \mu) \xrightarrow {d} N(0,\frac{\sigma^{2}}{n})$

$\bar{X}_{n} \xrightarrow {d} N(\mu,\frac{\sigma^{2}}{n})$

$ \sum X_{i} \xrightarrow {d} N(n\mu,\sigma^{2})$

Is it because the Slutsky theorem? (about continuous functions)

Thank's a lot!

  • 6
    $\begingroup$ None of those three statements, understood in its usual mathematical sense, makes any sense at all, because the limiting procedures implied by the arrows render "$n$" meaningless. $\endgroup$ – whuber May 18 '17 at 19:08
  • $\begingroup$ @whuber thanks a lot! I just wrote an answer, i think is in the right direction right? $\endgroup$ – S. Cow May 18 '17 at 21:40

This is a wrong statement

$$(\bar{X}_{n} - \mu) \xrightarrow {d} N\left (0,\frac{\sigma^{2}}{n}\right)$$

because it is translated "the left-hand side converges to a normal random variable as $n$ goes to infinity. But as $n$ goes to infinity,the right hand side acquires zero variance and becomes a degenerate random variable, a constant, not a normal distribution. Let alone that we should write something like $$(\bar{X}_{n} - \mu) \xrightarrow {d} N\left (0,\frac{\sigma^{2}}{\lim_{n\to \infty}(n)}\right)$$

The following statement is correct (although notation is not universal)

$$(\bar{X}_{n} - \mu) \sim_{\text{approx}}N\left (0,\frac{\sigma^{2}} {n}\right),\;\;\; n<\infty$$

and how "close" to this normal random variable is it will depend on the sample size in combination with the properties of the distribution $X$ follows.


I was thinking about something:

Suppose we have a suficiently big $n$, so we can state that:

$\sqrt{n}(\bar{X}_{n} - \mu) \sim N(0,\sigma^{2})$

Fix n, and we can ask ourselves the next question:

How is $(\bar{X}_{n} - \mu)$ distributed?

Since is a linear transformation of a normal, it still a normal and the parameters are:

$E[\bar{X}_{n} - \mu] = 0$ and $Var(\bar{X}_{n} - \mu) = \frac{\sigma^{2}}{n}$

So finally: $(\bar{X}_{n} - \mu) \sim N(0,\frac{\sigma^{2}}{n})$

For the others random variables is straightforward.

  • 2
    $\begingroup$ The intuition is good, but the mathematics isn't right. A sequence of random variables typically approaches a limiting distribution like the Normal without ever being Normal for any $n$. The intuition is that for "sufficiently large" $n$, the variables will be "sufficiently close to" Normal. The answer to "How is $\bar X_n-\mu$ distributed?" is still worked out the same way, but the justification is totally different: since $\sqrt{n}(\bar X_n-\mu)$ has a mean of $0$ and variance of $\sigma^2$, $\bar X_n$ has a mean of $\mu$ and variance of $\sigma^2/n$ and is still approximately Normal. $\endgroup$ – whuber May 18 '17 at 21:46

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