Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I show the following formula is true?

$$ \lim_{N\to\infty} p_{\hat{\mu}}(x) = \delta(x-\mu) $$ when $$ p_{\hat{\mu}}(x) = \frac{\sqrt{N}}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac12\frac{N}{\sigma^2}(x-\mu)^2\right). $$

It involves the probability density function for the sample mean of independent, identically distributed, Gaussian random variables.

We have: $X_i \sim \mathcal{N}(\mu, \sigma^2)$. $$ \hat{\mu} = \frac{1}{N}[x_1 + x_2 + \ldots + x_N] $$ Then, it can be shown that the p.d.f. of $\hat{\mu}$ is: $$ p_{\hat{\mu}}(x) = \frac{\sqrt{N}}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac12\frac{N}{\sigma^2}(x-\mu)^2\right). $$ Then, the result is: $$ \lim_{N\to\infty} p_{\hat{\mu}}(x) = \delta(x-\mu) $$ Notice that I cannot apply the usual: $$ \lim_{x\to\infty}f(x)g(x) $$ since the limit $\lim_{x\to\infty}f(x)$ doesn't exist (a limit which returns infinity is one way in which it does not exist).

If I let $C = \frac{1}{\sqrt{2\pi\sigma^2}}$, $A = -\frac12 \frac{N}{\sigma^2}(x-\mu)^2$, then $$ \lim_{N\to\infty}p_{\hat{\mu}}(x) = \frac{C}{2A}\frac{1}{\sqrt{N}}\frac{1}{\exp(AN)} $$

share|improve this question

2 Answers 2

up vote 3 down vote accepted

This illustrates that $\delta$ is not a function, but a functional: it does not reside in the world of functions but in operators on functions. By definition, you have that $\int_{-\infty}^\infty f(x)\delta(x-\mu)=f(\mu)$ for all nice functions $f$ (continuous and compactly supported), so what you really want to show is $\lim_{n\rightarrow\infty} \int_{-\infty}^\infty p_\hat{\mu}(x)f(x)dx=f(\mu)$ for all compactly supported continuous functions $f$. This is not hard by doing a change of variables that puts $n$ inside $f(x)$ and exploiting continuity.

share|improve this answer
    
Hello, I am curious how to show $\lim_{N\to\infty}\int_{-\infty}^{\infty}p_{\hat{\mu}}(x)f(x)\,\mathrm{d}x = f(\mu)$. For instance, I actually inserted the expression for the pdf into the integral, but could not see how the $f(x)$ inside the integral becomes the same function $f(\cdot)$ but with a different argument $\mu$. –  jrand Feb 2 '13 at 1:50
    
$\lim_{N\to\infty} \int_{-\infty}^{\infty}\frac{\sqrt{N}}{\sqrt{2\pi\sigma^2}} \exp(-\frac12 \frac{N}{\sigma^2}(x-2\mu x + \mu^2) f(x)\, \mathrm{d}x$ and I am wondering how to proceed. The goal is to somehow get the integral to result in the same function $f(\mu)$. ** I believe the purpose of doing this is to show that there exists a functional $\lim_{N\to\infty}p_{\hat{\mu}}(x)$ which exhibits the same properties as the delta functional. Then, the limit of the function is the same as the delta functional. –  jrand Feb 2 '13 at 1:53
    
@jrand: First, shift the integral via $x\rightarrow x-\mu$ to get rid of the $\mu$. Then let $u=\sqrt{N}x$. Then use the dominated convergence theorem. –  Alex R. Feb 2 '13 at 2:03
    
We wish to show $\int_{-\infty}^{+\infty}f(x)g(x)\,\mathrm{d}x = f(\mu)\Leftrightarrow g(x) = \delta(x-\mu)$. If conditions exist for the dominated convergence theorem to be applicable, $\lim_{N\to +\infty}\int_{-\infty}^{+\infty}f(x)g(x)\,\mathrm{d}x = f(\mu) \Rightarrow \int_{-\infty}^{+\infty}f(x)\lim_{N\to +\infty}g(x)\,\mathrm{d}x = f(\mu) \Leftrightarrow \lim_{N\to +\infty}g(x) = \delta(x - \mu)$. –  jrand Mar 21 '13 at 15:56
    
$\int_{-\infty}^{+\infty} \sqrt{\frac{N}{2\pi\sigma^2}} \exp\left(-\frac12 \frac{N}{\sigma^2} (x-\mu)^2\right)f(x)\,\mathrm{d}x = \int_{-\infty}^{+\infty}\sqrt{\frac{N}{2\pi\sigma^2}} \exp\left(-\frac12 \frac{N}{\sigma^2}u^2\right)f(u + \mu)\,\mathrm{d}u = \int_{-\infty}^{+\infty}\sqrt{\frac{N}{2\pi\sigma^2}} \exp\left(-\frac{1}{2\sigma^2}x^2\right)f\left(\frac{x}{\sqrt{N}} + \mu\right)\,\frac{\mathrm{d}x}{\sqrt{N}} = \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2\sigma^2}x^2\right)f\left(\frac{x}{\sqrt{N}}+\mu\right)\, \mathrm{d}x$ –  jrand Mar 21 '13 at 15:57

Alternatively, what you want to prove is equivalent to showing that $\hat{\mu}$ converges in probability to $\mu$. That is, for every $\epsilon > 0$,

$$\lim_{n \rightarrow \infty}{P(|\hat{\mu}-\mu|>\epsilon) = 0}$$

This follows from the law of large numbers.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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