Let $\bar{X}_n = X_1 + \dots X_n$ where $X_i \sim N(0,1)$. We can easily verify that $\bar{X}_n \sim N(0, 1/n)$.

Thus $\text{Var}(\bar{X}_n) = 1/n$.

Let the density of $X \sim N(0,1)$ be denoted $\phi(x) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}}$.

Apparently we can write express $\bar{X}_n = \frac{1}{\sqrt{n}} Z$ where $Z \sim N(0,1)$ is the standard normal distribution.

My attempt so far is to note that $\phi(\bar{x}_n) = \frac{1}{\sqrt{2 \pi} \frac{1}{\sqrt{n}}} e^{-\frac{x^2}{2 \frac{1}{n}}}$

We can then define $Z=\frac{\bar{X}_n - \mu}{\sigma} = \frac{\bar{X}_n}{\frac{1}{\sqrt{n}}}$ as usual. Then be substiution we obtain:

$\phi(\bar{x}_n) = \frac{1}{\frac{1}{\sqrt{n}}} \phi(z) = \sqrt{n} \phi(z)$

Why do I keep getting the factor of $\sqrt{n}$ on the numerator rather than the denominator?



You denote the prob. density by $\phi(x)$, I prefer $f_X(x)$.

If you like to rewrite this in terms of a new random variable $z$ you have to use a transformation, i.e. $$f_X(x) = f_Y(y) \left| \frac{dx}{dy}\right|$$ This is valid for continuous random variables, not for discrete random variables.


You have several errors caused by abuse of notations. First, let's define sample mean correctly: $$\bar{X_n}=\frac{1}{n}\sum_{i=1}^n X_i$$ Variance and mean are correct. The density of $\bar{X_n}$ can be correctly written as $p_{\bar{X_n}}(x)$ (or $\phi_{\bar{X_n}}(x)$ if you want, but I'll reserve $\phi$ for standard normal distribution). Your notation uses $\phi(\bar{x_n})$, as if you replace $x$ by $\bar{x_n}$ in the definition of $\phi(x)$. But, if we replace $\phi(\bar{x_n})$ by $p_{\bar{X_n}}(x)$, your density function for $\bar{X_n} $is correct, i.e. $$p_{\bar{X_n}}(x)=\frac{\sqrt{n}}{\sqrt{2\pi}}e^{-\frac{nx^2}{2}}$$ This can be written in terms of $\phi(x)$: $$p_{\bar{X_n}}(x)=\sqrt{n}(\sqrt{2\pi})^{n-1}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\right)^n=\sqrt{n}(2\pi)^{(n-1)/2}\phi(x)^n$$

Another comment on how you're trying to relate the two: $\sqrt{n}\phi(z)$ is not equal to $\phi(\bar{x_n})$; first of all the variables are different, and if they are PDFs, then their integral should be $1$, but due to $\sqrt{n}$ one of them is not.

  • $\begingroup$ Thanks. I follow your result that the x pdf can be written in terms of $\phi(x)^n$ but how does this show $\bqr{X}_n = n^{-1/2} Z$? See second to last line of page 1 in these notes for the claim: www.stat.cmu.edu/~larry/=stat705/Lecture2.pdf $\endgroup$ – user11128 Feb 24 '19 at 10:53
  • $\begingroup$ This doesn't show $\bar{X_n}=\frac{1}{\sqrt{n}}Z$, but I tried to answer your title question. For this one, in the notes, $\bar{X_n}$ is defined that way via $\overset{d}=$ operator. So, it's basically saying that if $\bar{X_n}$ is Normal with $(0,1/n)$, then a standard normal RV can be defined such that $\bar{X_n}=\frac{1}{\sqrt{n}}Z$. $\endgroup$ – gunes Feb 24 '19 at 11:10
  • $\begingroup$ Basically just applying Var(cX) = c^2 Var(X)? $\endgroup$ – user11128 Feb 24 '19 at 11:13
  • $\begingroup$ Yes, and it also applies the fact that normal RVs are still normal under linear transformation. Other RVs may not satisfy this property. For example, if $X$ is Bernoulli, $2X$ is not Bernoulli. $\endgroup$ – gunes Feb 24 '19 at 11:16

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