Probability question: random variables

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The discussion addresses two probability questions involving random variables. For the first question, it calculates the probability P(Y = 4) as 1/16 and the expected time E[Y] for the first HT occurrence as 2, with similar expected intervals for HT and TT also being 2. The second question focuses on estimating a random variable U based on noisy measurements Y1 and Y2, determining the linear mean-squared error estimate as 0.25Y1 + 0.75Y2. The minimum mean-squared error is calculated to be 5, considering the variances and covariances involved. Overall, the thread provides detailed solutions to both probability scenarios.
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I have two questions.. any insight into either of them is appreciated.

1) A fair coin is tossed repeatedly; the sequence of outcomes is recorded. Let Y be the random time of the first appearance of HT (a tail immediately following a head).
(a) Find P(Y = 4):
(b) Find E[Y] (the expected time to wait for the first HT).
(c) Find the expected length of the interval between successive appearances of HT.
(d) Same as (c) for TT.


2) U is a random variable having mean 2 and variance 5. Two noisy measurements of U are taken:
Y1 = U + Z1
Y2 = U + Z2:
where Z1; Z2; U are assumed pairwise uncorrelated, and where E[Z1] = E[Z2] = 0; Var(Z1) = 1; Var(Z2) = 2:
(a) Determine the linear mean-squared error (MMSE) estimate of U based on Y1 and Y2:
(b) Compute the resulting minimum mean-squared error.


Thank you.
 
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1) (a) P(Y = 4) = (1/2)^4 = 1/16, since the sequence of outcomes must be HTHH for Y = 4.
(b) E[Y] = 2, since the expected time to wait for HT is the same as the expected time to get a head, which is 2 tosses.
(c) The expected length of the interval between successive appearances of HT is also 2, since the probability of getting HT on any given toss is 1/2 and the sequence of outcomes is independent.
(d) Similarly, the expected length of the interval between successive appearances of TT is also 2, since the probability of getting TT on any given toss is 1/4 and the sequence of outcomes is independent.

2) (a) The linear mean-squared error estimate of U based on Y1 and Y2 is given by:
E[U|Y1,Y2] = aY1 + bY2, where a and b are constants.
To find these constants, we can use the fact that E[U|Y1,Y2] = U since U is the true value of U, and E[aY1 + bY2] = aE[Y1] + bE[Y2], where E[Y1] = E[Y2] = 2 (since E = 2) and E = 2.
Therefore, a + b = 1 and 2a + 2b = 2, which gives us a = 0.25 and b = 0.75.
So, the linear mean-squared error estimate of U based on Y1 and Y2 is 0.25Y1 + 0.75Y2.
(b) The resulting minimum mean-squared error is given by:
E[(U - E[U|Y1,Y2])^2] = Var(U) - Cov(U,Y1+Y2)^2/Var(Y1+Y2), where Var(U) = 5, Cov(U,Y1+Y2) = 0 (since U and Y1+Y2 are uncorrelated) and Var(Y1+Y2) = Var(Y1) + Var(Y2) = 1 + 2 = 3.
Therefore, the minimum mean-squared error is 5 - 0/3 = 5.
 
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