The data for the joint probability mass function of X and Y (two different measu

pavitorj6 2021-11-18 Answered
The data for the joint probability mass function of X and Y (two different measurement systems) are given in the table below.
a) Calculate the marginal distributions of X and Y and plot them.
b) Select one of the Y values from the table and find the conditional probability mass function of X for that Y value you have selected and plot it.
c) Show whether X and Y are independent or not.
\begin{array}{|c|c|}\hline f(x,y) & 1 & 2 & 3 & 4 \\ \hline 15 & 0.1 & 0 & 0.1 & 0.05 \\ \hline 20 & 0.05 & 0.05 & 0 & 0.1 \\ \hline 25 & 0 & 0.05 & 0.05 & 0.1 \\ \hline 30 & 0.1 & 0.05 & 0.15 & 0.05 \\ \hline \end{array}
d) Calculate the covariance of (X,Y) i.e. Cov(X,Y).

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Expert Answer

Alrew1959
Answered 2021-11-19 Author has 836 answers

Step 1
Given:
\[\begin{array}{|c|c|}\hline f(x,y) & 1 & 2 & 3 & 4 & \text{Total}\\ \hline 15 & 0.1 & 0 & 0.1 & 0.05 & 0.25 \\ \hline 20 & 0.05 & 0.05 & 0 & 0.1 & 0.2 \\ \hline 25 & 0 & 0.05 & 0.05 & 0.1 & 0.2\\ \hline 30 & 0.1 & 0.05 & 0.15 & 0.05 & 0.35 \\ \hline \text{total} & 0.25 & 0.15 & 0.3 & 0.3 & 1\\ \hline \end{array}\]
Step 2
a) Marginal distribution of X:
\[\begin{array}{|c|c|}\hline X & P(X) \\ \hline 15 & 0.25 \\ \hline 20 & 0.2 \\ \hline 25 & 0.2 \\ \hline 30 & 0.35 \\ \hline \text{total} & 1 \\ \hline \end{array}\]
image

\[\begin{array}{|c|c|}\hline X & P(Y) \\ \hline 1 & 0.25 \\ \hline 2 & 0.15 \\ \hline 3 & 0.3 \\ \hline 4 & 0.3 \\ \hline \text{total} & 1 \\ \hline \end{array}\]
image Step 3
b) Let \(\displaystyle{Y}={2}\)
\(\displaystyle{P}{\left({X}{\mid}{Y}={2}\right)}={\frac{{{P}{\left({X},{Y}={2}\right)}}}{{{P}{\left({Y}={2}\right)}}}}\)
Probability distribution of \(\displaystyle{X}{\mid}{Y}={2}\)
\[\begin{array}{|c|c|}\hline X & P(X|Y=2) \\ \hline 15 & 0 \\ \hline 20 & 0.333333 \\ \hline 25 & 0.333333 \\ \hline 30 & 0.333333 \\ \hline \text{total} & 1 \\ \hline \end{array}\]
Step 4
c) For X and Y to be independent, product of marginal probabilities should be equal to joint probability for all values of X and Y
\(\displaystyle{P}{\left({x}={15}\right)}={0.25}\)
\(\displaystyle{P}{\left({Y}={1}\right)}={0.25}\)
\(\displaystyle{P}{\left({x}={15},{y}={1}\right)}={01}\)
Here, \(\displaystyle{0.25}\cdot{0.25}={0.0625}\) (which is not equal to 0.1)
Thus , X and Y are not independent of each other.
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