On coin-tossing measure

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Suppose that $\{X_i\}$ are i.i.d r.vs with $P(X_i=0)=p, P(X_i=1)=1-p, p\in (0,1).$ Let $X=\sum_{n=1}^\infty\frac{X_n}{2^n}=\sum_{n=1}^\infty Y_n$ and $S_n=Y_1+\cdots+Y_n$. Then $P(Y_n=0)=p, P(Y_n=\frac{1}{2^n})=1-p.$ Moreover, $Y_n$ has distribution $p\delta_0+(1-p)\delta_{1/2^n}$. Therefore, $S_n$ has distribution $\mu_n=\ast_{k=1}^n(p\delta_0+(1-p)\delta_{1/2^k})$, where $\ast$ denotes convolution. We call the weak limit of $\mu_n$, say $mu=\ast_{n=1}^\infty(p\delta_0+(1-p)\delta_{1/2^n})$, the coin-tossing measure. (We can use characteristic function to prove $S_n$  converges weakly.)

When $p=1/2$, $\mu$ is the one dimensional Lebesgue measure. 

For example, $\mu_2([1/2^2,1/2))=P(S_2\in [1/2^2,1/2))=P(Y_1+Y_2\in [1/2^2,1/2))=P(Y_1=1/2,Y_2=0)=(1-p)p.$

Reference:Cosine products, Fourier transformations and random sums by Kent E. Morrison.

On coin-tossing measure

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原文地址：http://www.cnblogs.com/jinjun/p/4996503.html