2
$\begingroup$

My introductory probability book gives the following two theorems:

Theorem 1.

For large $n$, the distribution of $\bar{X}_n$ is approximately $N(\mu, \sigma^2/n)$.

Theorem 2.

The CLT says that the sample mean $\bar{X}_n$ is approximately Normal, but since the sum $W_n = X_1 + ... + X_n = n\bar{X}_n$ is just a scaled version of $\bar{X}_n$, the CLT also implies $W_n$ is approximately Normal. If the $X_j$ have mean $\mu$ and variance $\sigma^2$, $W_n$ has mean $n\mu$ and variance $n\sigma^2$. The CLT then states that for large $n$,

$W_n \sim N(n\mu, n\sigma^2)$.

It seems to me like the theorem 2 is inconsistent with theorem 1. If $W_n = n\bar{X}_n$, meaning it is just a scaled version of the sample mean, and the sample mean itself is approximately Normal, then, according to theorem 1, for large $n$, the distribution of $W_n = n\bar{X}_n$ should also approximately be $N(\mu, \sigma^2/n)$, no?

I would greatly appreciate it if people could please take the time to clarify this.

$\endgroup$
1
$\begingroup$

$$\Bbb{E}(W_n) = \Bbb{E}(n \bar{X}_n) = n \Bbb{E}(\bar{X}_n) \stackrel{\text{Thm 1}}{\approx} n \mu \\ \mathrm{var}(W_n) = \mathrm{var}(n \bar{X}_n) = n^2 \mathrm{var}(\bar{X}_n) \stackrel{\text{Thm 1}}{\approx} n^2 \cdot \frac{\sigma^2}{n} = n \sigma^2$$

$\endgroup$
3
$\begingroup$

$W_n$ is the sample total, namely $$W_n = \sum_{i=1}^n X_i,$$ where $X_i$ are IID random variables with mean $\mu$ and standard deviation $\sigma$. Then by linearity of expectation, we have exactly $$\operatorname{E}[W_n] = \operatorname{E}\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n \operatorname{E}[X_i] = \sum_{i=1}^n \mu = n\mu.$$ Furthermore, because the $X_i$s are independent, the variance of the sum equals the sum of the variances: $$\operatorname{Var}[W_n] = \operatorname{Var}\left[\sum_{i=1}^n X_i\right] \overset{\text{ind}}{=} \sum_{i=1}^n \operatorname{Var}[X_i] = \sum_{i=1}^n \sigma^2 = n\sigma^2.$$ So even before invoking any laws of large samples or assumptions about the parametric family of $\{X_i\}_{i = 1}^n$, we have exact relationships for the mean and variance of the sample total; and in turn, we also have exact relationships for the first two moments of $\bar X_n = W_n/n$. The only condition we needed is that the sample is IID and that the first two moments exist.

Now, any large sample theory must be consistent with the above. So Theorem 1 states $$\bar X_n \overset{\approx}{\sim} \operatorname{Normal}(\mu,\sigma^2/n),$$ thus $W_n = nX_n$ is also approximately $n$ times such a normal random variable. When we multiply a normal random variable by a scalar $n$, we multiply its mean by $n$ and its variance by $n^2$, which again is consistent with the above.

As a concrete example, suppose $X \sim \operatorname{Binomial}(n = 5, p = 1/3)$. Then $$\operatorname{E}[X] = np = 5/3 = \mu, \quad \operatorname{Var}[X] = np(1-p) = 10/9 = \sigma^2.$$ Now say I take a sample of such $X$ of size $n^* = 90$ (where I use $n^*$ to distinguish this sample size from the binomial parameter $n$). If I take the total $W_{90}$, you could reason that because each $X_i$ is on average $5/3$, the total should be on average $90(5/3) = 150$. It would be absurd to think that $W_{90}$, the total of $90$ such binomial random variables, should have the same mean as a single $X$, which is what you essentially claimed by invoking Theorem $1$ on $W_n$. Your intuition should contradict this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.