Calculating the distribution of the random variable Y at the outputThe random variable X at the entrance to the transmission path is sufficient for the following distribution. X = P ( x ) x 1 = 0.1 x 2 = 0.2 x 3 = 0.3 x 4 = 0.4The random variable Y at the output of the transmission link only takes three values. The transfer matrix T with p i j = P ( Y = y j | X = x i ) is T = ( p i j ) = ( 0.50 0 0.50 0.20 0.40 0.40 0.50 0.25 0.25 0 0.50 0.50 )

Question

Calculating the distribution of the random variable Y at the outputThe random variable X at the entrance to the transmission path is sufficient for the following distribution.  X  =  P  (  x  )      x    1    =  0.1       x    2    =  0.2       x    3    =  0.3       x    4    =  0.4The random variable Y at the output of the transmission link only takes three values. The transfer matrix T with       p          i              j        =  P  (  Y  =      y    j        |    X  =      x    i    ) is  T  =  (      p          i              j        )  =      (                            0.50                          0                          0.50                                      0.20                          0.40                          0.40                                      0.50                          0.25                          0.25                                      0                          0.50                          0.50                      )

last99erib · Accepted Answer

Step 1What you have is a matrix of conditional probabilities, with the entry in row i, column j of T corresponding to   P  (  Y  =      y    j        |    X  =      x    i    ). To find the marginal distribution of Y, use the law of total probability:  P  (  Y  =      y    j    )  =      ∑          i      =      1        4    P  (  Y  =      y    j        |    X  =      x    i    )  P  (  X  =      x    i    )Step 2Note: If p represents a   1  ×  4 row vector summarizing the probability masses of X, then pT is a   1  ×  3 row vector summarizing the masses of Y.