Let n be a natural number and let S=\{0,1\}^{n}. Recall that given p with 0<p<1 fixed....

[FreddyB]

Let $n$ be a natural number and let $S=\{0,1\}^{n}$. Recall that given $p$ with $0<p<1$ fixed, the binomial measure $\beta: \wp(S) \rightarrow[0,1]$ is determined by

$\qquad \qquad \beta\left(\left\{\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right)\right\}\right)=p^{\#\left\{j: \omega_{j}=1\right\}}(1-p)^{\#\left\{j: \omega_{j}=0\right\}}$

(a) Express the binomial measure as a product measure in terms of the probability measure $\pi: \wp(\{0,1\}) \rightarrow[0,1]$ with $\pi(\{1\})=p$.

(b) Taking $n=3$ consider the sets

$\qquad \qquad x^{-1}(\{j\}) \quad \text { for } j=0,1,2,3$

where $x: S \rightarrow \mathbb{R}$ by

$\qquad \qquad x\left(\omega_{1}, \omega_{2}, \omega_{3}\right)=\omega_{1}+\omega_{2}+\omega_{3}$

(i) Find $x^{-1}(\{j\})$ for $j=0,1,2,3$.

(ii) Taking $p=1 / 2$, find $M(j)=\alpha(\{j\})=\beta\left(x^{-1}(\{j\})\right)$ for $j=0,1,2,3$.

(iii) Taking $p=3 / 4$, find $M(j)=\alpha(\{j\})=\beta\left(x^{-1}(\{j\})\right)$ for $j=0,1,2,3$.

(c) Generalize/repeat part (b) for $n=4,5,6$

(d) Compute the integral of $x$ with respect to the binomial measure $\beta$ (for general $n$ and $p$ ).

 

[stapel]

Let $n$ be a natural number and let $S=\{0,1\}^{n}$. Recall that given $p$ with $0<p<1$ fixed, the binomial measure $\beta: \wp(S) \rightarrow[0,1]$ is determined by

$\qquad \qquad \beta\left(\left\{\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right)\right\}\right)=p^{\#\left\{j: \omega_{j}=1\right\}}(1-p)^{\#\left\{j: \omega_{j}=0\right\}}$

(a) Express the binomial measure as a product measure in terms of the probability measure $\pi: \wp(\{0,1\}) \rightarrow[0,1]$ with $\pi(\{1\})=p$.

(b) Taking $n=3$ consider the sets

$\qquad \qquad x^{-1}(\{j\}) \quad \text { for } j=0,1,2,3$

where $x: S \rightarrow \mathbb{R}$ by

$\qquad \qquad x\left(\omega_{1}, \omega_{2}, \omega_{3}\right)=\omega_{1}+\omega_{2}+\omega_{3}$

(i) Find $x^{-1}(\{j\})$ for $j=0,1,2,3$.

(ii) Taking $p=1 / 2$, find $M(j)=\alpha(\{j\})=\beta\left(x^{-1}(\{j\})\right)$ for $j=0,1,2,3$.

(iii) Taking $p=3 / 4$, find $M(j)=\alpha(\{j\})=\beta\left(x^{-1}(\{j\})\right)$ for $j=0,1,2,3$.

(c) Generalize/repeat part (b) for $n=4,5,6$

(d) Compute the integral of $x$ with respect to the binomial measure $\beta$ (for general $n$ and $p$ ).

Please reply with a clear listing of your thoughts and efforts so far, so that we can begin working with you. (Read Before Posting)

Thank you!

 

[FreddyB]

I have included my current efforts as well as what I am unsure how to complete the problem below.

a) To express the binomial measure as a product measure in terms of the probability measure π defined on {0,1}, I express each individual set {ωj} in terms of π and then use this to obtain an expression for β(A) as a product over j.

b) i) For j = 0, the only combination is (0, 0, 0). For j = 1, there are three combinations: (0, 0, 1), (0, 1, 0), and (1, 0, 0). For j = 2 and j = 3 there are also three combinations each. Therefore the pre-images are: x^-1(0) = (0,0,0), x^-1(1) = (1,0,0), (0,1,0), (0,0,1), x^-1(2) = (1,1,0), (1,0,1), (0,1,1), x^-1(3) = (1,1,1)

ii) Unsure how to do

c) Unsure how to do

d) The integral of x with respect to the binomial measure β is computed by summing the product of x and β(x) over all x in the domain. This can be simplified by first computing the binomial measure for each n-tuple, then using these measures to compute the integral. The final answer is ((1 - p)^(Count[j:x_j = 0]) p^(Count[j:x_j = 1]))^n.