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  • Limiting Distributions

Convergence in Distribution

• Can be the weakest Convergence, but can be hard to prove.
• $F_X(x)$ is our Cumulative Distribution Function
  • $F_X(x)=P(X\le x)$
• If I have $X_1\sim F_{X_1}(x),X_2\sim F_{X_2}(x),\dots$
• If we have $X\sim F(x)$
  • $\underset{n\to \mathrm{\infty }}{lim}{F}_{X_n}(x)=F_x$
  • $X_n\stackrel{d}{\to }X$
• Slutsky's Theorem #tk
• This can be proven using the Moment Generating Function.
• Theorem 6
  • Let $\{X_n\}$ be a sequences of random variables iwht mgf $M_{X_n}(t)=M(t)$ for $\left|t\right|\le h$ then $X_n\stackrel{d}{\to }X$
• Example:
  • What does the Binomial Distribution converge in distribution to?
  • $X_n$ is a sequence of Random Variables
  • $\mathrm{Bin}(n,p)$
  • Suppose mean is $\mu =np$
• Example: Slide 14
  • MGF of a Binomial Distribution
  • $M_X(t)=[p{e}^t+(1-p){]}^n$
  • $e^x=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{x}{n})}^n$
  • $e^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^0}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\dots$
  • #tk remember these
  • $X_n\sim \mathrm{Bin}(n,p)$
  • $p=\frac{\mu }{n}$
  • $np=\mu$
  • $X_1\sim \mathrm{Bin}(1,p)$
    • Bernoulli Distribution
  • $X_2\sim \mathrm{Bin}(2,p)$
  • $X_3\sim \mathrm{Bin}(3,\mathrm{p})$
  • $\dots$
  • $X_n\sim \mathrm{Bin}(n,p)$
  • The MGF of $X_n$ is $M_{X_n}(t)=[p{e}^t+(1-p){]}^n$
  • Since $p=\frac{\mu }{n}$
  • $M_{X_n}(t)={[\frac{\mu }{n}{e}^t+(1-\frac{\mu }{n})]}^n$
  • $M_{X_n}(t)={[\frac{\mu }{n}{e}^t+1-\frac{\mu }{n}]}^n$
  • $M_{X_n}(t)={[1-\frac{\mu }{n}(1-e^t)]}^n$
  • Find what it converges to in distribution
  • $\underset{n\to \mathrm{\infty }}{lim}{M}_{X_n}(t)=\underset{n\to \mathrm{\infty }}{lim}{[1-\frac{\mu }{n}(1-e^t)]}^n$
  • $\underset{n\to \mathrm{\infty }}{lim}{M}_{X_n}(t)=\underset{n\to \mathrm{\infty }}{lim}{[1+\frac{\mu }{n}(-1+e^t)]}^n$
  • $\underset{n\to \mathrm{\infty }}{lim}{M}_{X_n}(t)=\underset{n\to \mathrm{\infty }}{lim}{[1+\frac{\mu }{n}(e^t-1)]}^n$
  • Since $e^x=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{x}{n})}^n$
  • let $x=e^t-1$
  • $\underset{n\to \mathrm{\infty }}{lim}{M}_{X_n}(t)=\underset{n\to \mathrm{\infty }}{lim}{[1+\frac{\mu }{n}(x)]}^n$
  • So we have $e^{\mu x}=\underset{n\to \mathrm{\infty }}{lim}{[1+\frac{\mu x}{n}]}^n$
  • $e^{\mu (e^t-1)}=\underset{n\to \mathrm{\infty }}{lim}{[1+\frac{\mu (e^t-1)}{n}]}^n$
  • So $\underset{n\to \mathrm{\infty }}{lim}{M}_{X_n}(t)=\underset{\text{MGF of poisson}}{\underbrace{e^{\mu (e^t-1)}}}$
  • This is Poisson Distribution, basically the derivation proof.
  • See that if $X\sim \mathrm{Poisson}(\lambda )$
    • The PDF is $f(x)=\frac{e^{-\lambda }{\lambda }^x}{x!}x=0,1,2,\dots$
    • To derive the MGF, we need to do $M_X(t)=\mathrm{E}[e^{tx}]$
    • $M_X(t)=\sum _{i=0}^{\mathrm{\infty }}{e}^{t{x}_i}f(x_i)$
    • $M_X(t)=\sum _{x=0}^{\mathrm{\infty }}{e}^{tx}f(x)$
    • $M_X(t)=\sum _{x=0}^{\mathrm{\infty }}{e}^{tx}\frac{e^{-\lambda }{\lambda }^x}{x!}$
    • $M_X(t)=e^{-\lambda }\sum _{x=0}^{\mathrm{\infty }}{e}^{tx}\frac{{\lambda }^x}{x!}$
    • $M_X(t)=e^{-\lambda }\sum _{x=0}^{\mathrm{\infty }}\frac{e^{tx}{\lambda }^x}{x!}$
    • $M_X(t)=e^{-\lambda }\sum _{x=0}^{\mathrm{\infty }}{\left(\frac{\lambda e^t}{x!}\right)}^x$
      • This is the taylor series expansion of $e^x$
    • $e^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\dots$
    • $M_X(t)=e^{-\lambda }\cdot e^{\lambda e^t}$
    • $M_X(t)=e^{\lambda (e^t-1)}$
    • This is the MGF of the Poisson Distribution