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

[FreddyB]

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

[imath]\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\}}[/imath]

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

(b) Taking [imath]n=3[/imath] consider the sets

[imath]\qquad \qquad x^{-1}(\{j\}) \quad \text { for } j=0,1,2,3[/imath]

where [imath]x: S \rightarrow \mathbb{R}[/imath] by

[imath]\qquad \qquad x\left(\omega_{1}, \omega_{2}, \omega_{3}\right)=\omega_{1}+\omega_{2}+\omega_{3} [/imath]

(i) Find [imath]x^{-1}(\{j\})[/imath] for [imath]j=0,1,2,3[/imath].

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

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

(c) Generalize/repeat part (b) for [imath]n=4,5,6[/imath]

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

 

[stapel]

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

[imath]\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\}}[/imath]

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

(b) Taking [imath]n=3[/imath] consider the sets

[imath]\qquad \qquad x^{-1}(\{j\}) \quad \text { for } j=0,1,2,3[/imath]

where [imath]x: S \rightarrow \mathbb{R}[/imath] by

[imath]\qquad \qquad x\left(\omega_{1}, \omega_{2}, \omega_{3}\right)=\omega_{1}+\omega_{2}+\omega_{3} [/imath]

(i) Find [imath]x^{-1}(\{j\})[/imath] for [imath]j=0,1,2,3[/imath].

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

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

(c) Generalize/repeat part (b) for [imath]n=4,5,6[/imath]

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

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.