April Bush

Answered

2022-06-27

Calculating the distribution of the random variable Y at the output

The random variable X at the entrance to the transmission path is sufficient for the following distribution.

$X=P(x)$

${x}_{1}=0.1$ ${x}_{2}=0.2$ ${x}_{3}=0.3$ ${x}_{4}=0.4$

The 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

$T=({p}_{i}^{j})=\left(\begin{array}{ccc}0.50& 0& 0.50\\ 0.20& 0.40& 0.40\\ 0.50& 0.25& 0.25\\ 0& 0.50& 0.50\end{array}\right)$

The random variable X at the entrance to the transmission path is sufficient for the following distribution.

$X=P(x)$

${x}_{1}=0.1$ ${x}_{2}=0.2$ ${x}_{3}=0.3$ ${x}_{4}=0.4$

The 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

$T=({p}_{i}^{j})=\left(\begin{array}{ccc}0.50& 0& 0.50\\ 0.20& 0.40& 0.40\\ 0.50& 0.25& 0.25\\ 0& 0.50& 0.50\end{array}\right)$

Answer & Explanation

last99erib

Expert

2022-06-28Added 19 answers

Step 1

What 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:

$P(Y={y}_{j})=\sum _{i=1}^{4}P(Y={y}_{j}|X={x}_{i})P(X={x}_{i})$

Step 2

Note: If p represents a $1\times 4$ row vector summarizing the probability masses of X, then pT is a $1\times 3$ row vector summarizing the masses of Y.

What 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:

$P(Y={y}_{j})=\sum _{i=1}^{4}P(Y={y}_{j}|X={x}_{i})P(X={x}_{i})$

Step 2

Note: If p represents a $1\times 4$ row vector summarizing the probability masses of X, then pT is a $1\times 3$ row vector summarizing the masses of Y.

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