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The question is based on the following problem in the book Probability Essentials by Jacod: enter image description here

Here is my question:

What does $P(\{k\})$ mean in the second problem?

The probability on a finite or countable measurable space $(\Omega,\Sigma)$ is determined by $P(\{\omega\})$ where $\omega\in\Omega$. As I understand, for the binomial distribution $B(p,n)$ the sample space is $\Omega=\{(a_1, a_2):a_1,a_2=0,1\}^n$. How does $P(\{k\})$ come out here?

4 Answers 4

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It is an abuse of notation. If we denote the probability space as $(\Omega,\Sigma,P)$, and let $X:\Omega\to{\Bbb R}$ be a random variable with Binomial distribution $B(p,n)$.

$P$ in $P(\{k\})$ should be understood as the distribution measure of $X$ and $P(\{k\})$ means $P(X=k)$, which is defined as $P(\{\omega\in\Omega:X(\omega)=k\})$.

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$k$ is the number of success which is an event $\in\Sigma$. Namely trying $n$ times getting the success of $k$ times. Therefore this event has some probability $P(\{k\})$. I also think that it is defined in $4.1$ and there is an abuse of notation.

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$P(\{k\})$ in Exercise 4.2 is $P(k\ \text{successes})$ in Exercise 4.1.

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    @Eric: Sure, with $\{k\}$ identified to $[k\ \text{successes}]$.2012-09-06
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The binomial random variable is the number of success in n trials. It can take on any integer value from 0 to n and the probability that a binomial with success proportion p and n trials equals k (i.e. there are exactly k successes in the n trials) is C$_k$^n$ p$^k$ (1-p)$^n$^-$^k$ where C$_k$^n$ is the number of ways to pick k objects out of a total of n objects. So P({k}) is just this probability of getting exactly k successes in n trials.