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I'm currently studying this proof

In this proof the author defines a probability measure

$$P^*[\{\omega\}]=(p^*)^{k(\omega)}(1-p^*)^{T-k(\omega)}$$ on $$\Omega=\{\omega=(y_1,\ldots,y_T)|y_i=\pm1\}$$

where $p^*=(r-a)/(b-a)$ and $k(\omega)$ is the number of ones in $\omega$.

$a<r<b$.

Unfortunately I can't prove that $P^*$ is indeed a probability measure.

$P^*[\{\omega\}]\ge0$ is clear.

I don't see $P^*(\Omega)=1$

$P^*(\Omega)=\sum_{\omega_i \in \Omega}(p^*)^{k(\omega_i)}(1-p^*)^{T-k(\omega_i)}$, but I don't know how to continue

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If you consider $\omega$ and $\tilde{\omega}$ with $k(\omega)=k(\tilde{\omega})$ it holds that $P^*(\omega)=P^*(\tilde{\omega})$. Now instead of summing up over every $\omega_i \in \Omega$ you can sum up from $n =0 ... T$ and count the elements with $k(\omega_i)= n$. There are $\dbinom{T}{n}$ elements in $\Omega$ which fullfill $k(\omega_i)= n$.

Therefore $P^*(\Omega)=\sum_{\omega_i \in \Omega}(p^*)^{k(\omega_i)}(1-p^*)^{T-k(\omega_i)} = \sum_{n=0}^T \dbinom{T}{n} (p^*)^n(1-p^*)^{T-n}=1$ as a result from the binomial theorem.

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  • $\begingroup$ Thanks a lot, that's a clever way. How would you show additivity? $\endgroup$ – user674879 May 17 at 16:40

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