Question

5.2.6 An article in Health Economics [“Estimation of the Transi- tion Matrix of a Discrete-Time Markov Chain” (2002, Vol. 11, pp. 33–42)] considered the changes in CD4 white blood cell counts from one month to the next. The CD4 count is an important clini- cal measure to determine the severity of HIV infections. The CD4 count was grouped into three distinct categories: 0–49, 50–74, and > 75. Let X and Y denote the category minimum) CD4 count at a month and the following month, respectively. The conditional probabilities for Y given values for X were provided by a transition probability matrix shown in the following table. 75 0.0059 0 0.9819 0.1766 0.0237 50 0.0122 0.7517 0.0933 0.0717 0.8830

Answer 1

It is given that, the transition probability matrix of conditional probability for given

and the probability distribution of is
P(X = 75)=0.9,P(X = 50)=0.08
and
P(X = 0) = 0.02
.

P(Y = 50, X = 50) P(X = 50) P(Y S50|X = 50) = P(Y = 0, X = 50) P(X = 50) = 0.1766+0.7517 = 0.9283

Therefore, the value of
P(Y <50 X = 50)
is
10.92831

P(X = 0, y = 75) = P(Y = 75 X = 0)*P( X = 0) = 0.0059(0.02) = 0.00012

Therefore, the value of
P(x = 0, y = 75)
is
[0.00012]

E(Y|X = 50) = { Y{ (Y|X) =0P(Y = 0,X = 50) +50P(Y = 50, X = 50) + 75P(Y = 75,X = 50) fo{P(Y = 0 | X = 50)~P(X = 50)}+50{P(Y = 50| X = 50)x P(X = 50) +75{P(Y = 75| X = 50)x P(X = 50)} = 0(0.1766x0.08) +50(0.7517x0.08)+75(0.0717x0.08) = 3.437

Therefore, .

E(Y|X = 50) = 3.437

Find the marginal distributions of .

First, find the joint probability density function of the random variables and.

For
X = 0,
:

P(Y =0|X = 0)=P(X = 0, y = 0) P(X =0) = P(x = 0, = 0) = P(Y =0|X = 0)*P(X =0) = 0.9819(0.02) = 0.019638

For
X = 0,

Y = 50
:

P(Y = 50| X = 0)=P(= 0, = 50) P(X =0) = P(X = 0, y = 50)=P(Y = 50| X =0)~P(X =0) = 0.0122(0.02) = 0.000244

For
X = 0,

Y = 75
:

P(Y = 75| X = 0) = P(X = 0,9 = 75) P(X=0) = P(X = 0,9 = 75) = P(Y = 75| X = 0)*P(X = 0) = 0.0059(0.02) = 0.00012

Similarly, the remaining values are presented in the below table:

X 50 75 Total 0 50 75 Total 0.01964 0.00024 0.00012 0.02 0.014130.06014 0.005740.08 0.02133 0.08397 0.7947 | 0.9 0.0551 0.14435 0.80055

From the above table,
P(Y = 75) = 0.80055
,
P(Y = 50) = 0.14435
and
P(Y=0) = 0.0551
.

.

fx (x,y)

The joint probability density function of the random variables and is shown below:

For
X = 0,
:

P(Y =0|X = 0)=P(X = 0, y = 0) P(X =0) = P(x = 0, = 0) = P(Y =0|X = 0)*P(X =0) = 0.9819(0.02) = 0.019638

For
X = 0,

Y = 50
:

P(Y = 50| X = 0)=P(= 0, = 50) P(X =0) = P(X = 0, y = 50)=P(Y = 50| X =0)~P(X =0) = 0.0122(0.02) = 0.000244

For
X = 0,

Y = 75
:

P(Y = 75| X = 0) = P(X = 0,9 = 75) P(X=0) = P(X = 0,9 = 75) = P(Y = 75| X = 0)*P(X = 0) = 0.0059(0.02) = 0.00012

Similarly, the remaining values are presented in the below table:

x O 50 75 Total у 0 50 75 0.019640.00024 0.00012 0.014130.06014 0.00574 0.02133 0.08397 0.7947 0.0551 0.14435 0.80055 Total 0.02 0.08 0.9 1

From the joint probability distribution table,

P( X = 75, Y = 75)=0.7947

P(X = 75)P(Y = 75) = 0.90x0.80055 = 0.720495

Here, it can be observed that

fx, y(x,y)#f2(x)f,(y)

Therefore, and are not independent.

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