# Sample Mean expressed using Standard Normal Distribution

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?

Thanks

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.
• 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 – user11128 Feb 24 '19 at 10:53
• 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 $\bar{X_n}=\frac{1}{\sqrt{n}}Z$. – gunes Feb 24 '19 at 11:10
• 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. – gunes Feb 24 '19 at 11:16