Question

Consider an image, in which every pixel takes a value of 1, with probability q, and a value 0, with probability 1−q, where q is the realized value of a random variable Q which is distributed uniformly over the interval [0,1]. The realized value q is the same for every pixel. Let Xi be the value of pixel i. We observe, for each pixel the value of Yi=Xi+N, where N is normal with mean 2 and unit variance. (Note that we have the same noise at each pixel.) Assume that, conditional on Q, the Xi are independent, and that the noise N is independent of Q and the Xi's.

Answer 1

Consider an image, in which every pixel takes a value of 1, with probability q, and a value 0, with probability1-q , where q is the realized value of a random variable Q which is distributed uniformly over the interval [0.1]. The realized value q is the same for every pixel.

Let Xi be the value of pixel i. We observe, for each pixel the value of Yi=Xi+N, where N is normal with mean 2 and unit variance. (Note that we have the same noise at each pixel.) Assume that, conditional on Q, the Xi's are independent, and that the noise N is independent of and then Xi's 's.

1. Find E[Yi]. (Give a numerical answer.)

2. Find Var[Yi]. (Give a numerical answer.)

3. Let be the event that the actual values X1 and X2 of pixels 1 and 2, respectively, are zero. Find the conditional probability of Q given A.
(Enter your answer in terms of in standard notation.)

For : 0<=q<=1, Find: fQ|A(q)=

Answer 2

Can anyone please help here?

Answer 3

1. should be 5/2.

Need help with #2 and #3

Answer 4

Ans for #2:
---------------
Var[Yi] = 13/12

Ans for #3:
----------------
fQ|A(q)= 3(1-q)^2

Answer 5

1.) 5/2
2.) 5/4
3.) 4(1-q)^2

Answer 6

Which one is the correct one?

Answer 7

how did you get 4(1-q)^2 ? can you please explain ?

Answer 8

3.) Bayes rule:
f_Q(q) * f_X1|Q(x1= 0 | q) * f_X2|Q(x2 = 0 | q) / (f_X1(x1=0) * f_X2(x2 = 0))

f_Q(q) = 1
f_X1|Q (x1=0|q) = 1-q
f_X2|Q (x2 =0|q) = 1-q
f_X1(x1=0) = integral from 0 to 1 of f_Q(q)*f_X1|Q(x1= 0 | q) dq (total probability theorem)
= integral from 0 to 1 of 1-q dq = 1/2
by the same token,
f_X2(x2=0) = 1/2
....
putting everything together
ans = 1*(1-q)*(1-q) / (1/4) = 4*(1-q)^2

Answer 9

thanks a ton

Answer 10

No, (3) ans should be 3(1-q)^2
the denominator should be (f_A) not (f_X1) * (f_X2)
f_A = (f_Q * f_A|Q * dq) integrated from 0 to 1
f_A = (1* (1-q)^2 * dq) integrated from 0 to 1, which is 1/3

Answer 11

1.) 5/2
2.) 5/4
3.) 3(1-q)^2