April Bush

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.
$X=P\left(x\right)$
${x}_{1}=0.1$ ${x}_{2}=0.2$ ${x}_{3}=0.3$ ${x}_{4}=0.4$
The random variable Y at the output of the transmission link only takes three values. The transfer matrix T with ${p}_{i}^{j}=P\left(Y={y}_{j}|X={x}_{i}\right)$ is
$T=\left({p}_{i}^{j}\right)=\left(\begin{array}{ccc}0.50& 0& 0.50\\ 0.20& 0.40& 0.40\\ 0.50& 0.25& 0.25\\ 0& 0.50& 0.50\end{array}\right)$

last99erib

Expert

Step 1
What you have is a matrix of conditional probabilities, with the entry in row i, column j of T corresponding to $P\left(Y={y}_{j}|X={x}_{i}\right)$. To find the marginal distribution of Y, use the law of total probability:
$P\left(Y={y}_{j}\right)=\sum _{i=1}^{4}P\left(Y={y}_{j}|X={x}_{i}\right)P\left(X={x}_{i}\right)$
Step 2
Note: 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.

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