Question

1. Suppose we are going to sample n individuals and ask each sampled person whether they support policy A or not. Let Yi Y0 otherwise 1 if person i in the sample supports the policy, and (a) Assume Y1, , Yn are, conditional on θ. 1.1.d. binary random variables with expec- tation θ. Write down the joint distribution Pr(Yi-yi, . . . ,Ý,-yn(9) in a compact form. Also write down the form of Pr(> Ý,-y|0) (b) For the moment, suppose you believed that θ e {0.0.0.1, , 0.9.1.0). Given that the results of the survey where Σ Y. = 62 for n = 100, compute Pr(ΣΥ, 629) for each of these 11 values of θ and plot these probabilities as a function of θ. (c) Now suppose you had no prior information to believe one of these θ-values over Bayes' rule to compute p(이 Σ Yi = 62) for each θ-value. Make a plot of this posterior distribution as a function of θ (d) Now suppose you allow θ to be any value in the interval [0,1]. Using the uniform prior density for θ, so that p(9)-1 , plot the posterior density p(9) x Pr(Σ Y,-62(9) as a function of θ

Answer 1

$(a)~P(Y_1=y_1,Y_2=y_2,...,Y_n=y_n|\theta)=\prod_{i=1}^nP(Y_i=y_i|\theta)\\ Since~Y_i's~conditional~on~\theta~are~independent~and~E(Y_i)=P(Y_i=1)=\theta\\ then~P(Y_1=y_1,Y_2=y_2,...,Y_n=y_n|\theta)=\prod_{i=1}^n\theta^{y_i}(1-\theta)^{1-y_i}\\=\theta^{\sum_{i=1}^ny_i}(1-\theta)^{n-\sum_{i=1}^ny_i}\\ P(\sum_{i=1}^nY_i=y|\theta)=\binom{n}{y}\theta^y(1-\theta)^{n-y};~y=0,1,2,...\\ =0~otherwise\\ (b)$

$Denote:~Y=\sum_{i=1}^nY_i~and~P(\sum_{i=1}^nY_i=62|\theta)=P(Y=62,\theta).$

R code:

theta=0:10*0.1
P=choose(100,62)*theta^(62)*(1-theta)^(100-62)
plot(theta,P,lwd=2,type="p",ylab=expression(P(Y==62,theta),xlab=expression(theta))

Output:

[1] 0.000000e+00 1.034671e-36 5.430828e-20 2.810801e-11 4.477688e-06
[6] 4.472880e-03 7.537790e-02 1.906659e-02 1.528642e-05 8.253201e-14
[11] 0.000000e+00

$(c)~P(\theta=x|\sum Y_i=62)=\frac{P(\sum_{i=1}^nY_i=62|\theta=x)p_{\theta}(x)}{\sum_{\theta}P(\sum_{i=1}^nY_i=62|\theta=x)p_\theta(x)}\\ since~p_\theta(0.0)=p_\theta(0.1)=p_\theta(0.2)=...=p_\theta(1.0)=1/11\\ P(\theta=x|\sum Y_i=62)=\frac{P(\sum_{i=1}^nY_i=62|\theta=x)}{\sum_{\theta}P(\sum_{i=1}^nY_i=62|\theta=x)}=P(\theta,~Y=62),~say\\$

R code:

theta=0:10*0.1
P=choose(100,62)*theta^(62)*(1-theta)^(100-62)
p1=P/(11*sum(P))
p1
plot(theta,p1,lwd=2,type="p",ylab=expression(p(theta)*P(theta,Y==62)),xlab=expression(theta))

Output:

[1] 0.000000e+00 9.507147e-37 4.990155e-20 2.582725e-11 4.114355e-06
[6] 4.109938e-03 6.926152e-02 1.751947e-02 1.404604e-05 7.583513e-14
[11] 0.000000e+00

(d)

R code:

theta=0:10*0.1
P=choose(100,62)*theta^(62)*(1-theta)^(100-62)
p1=P/(sum(P))
p1
plot(theta,p1,lwd=2,type="p",ylab=expression(p(theta)*P(theta,Y==62)),xlab=expression(theta))

Output:

[1] 0.000000e+00 1.045786e-35 5.489171e-19 2.840997e-10 4.525791e-05
[6] 4.520932e-02 7.618767e-01 1.927142e-01 1.545064e-04 8.341864e-13
[11] 0.000000e+00

(e)

R code:

theta=0:10*0.1
P=theta^(62)*(1-theta)^(38)*choose(100,62)*theta^(62)*(1-theta)^(100-62)
p1=P/(sum(P))
p1
plot(theta,p1,lwd=2,type="p",ylab=expression(p(theta)*P(theta,Y==62)),xlab=expression(theta))

Output:

p1
[1] 0.000000e+00 1.765010e-70 4.862671e-37 1.302576e-19 3.305601e-09
[6] 3.298506e-03 9.367653e-01 5.993614e-02 3.852605e-08 1.123020e-24
[11] 0.000000e+00