Question
Let the joint probability density function of two discrete random X and Y be given as:
| X | |||||
|---|---|---|---|---|---|
| 2 | 3 | 4 | 5 | ||
| Y | 0 | 0 | 0.03 | 0 | 0 |
| 1 | 0.34 | 0.30 | 0.16 | 0 | |
| 2 | 0 | 0 | 0.03 | 0.14 | |
ii) Find the marginal distribution of X and Y.
iii) Find the conditional distribution of X given .
iv) Test the independence of variable s X and Y.
Let , where
and
Find the distribution of:
v) Find .
Answer :
Word Count : 550
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Marginal distribution of X: [ P_X(2) = 0 + 0.34 + 0 = 0.34 ] [ P_X(3) = 0.03 + 0.30 + 0 = 0.33 ] [ P_X(4) = 0 + 0.16 + 0.03 = 0.19 ] [ P_X(5) = 0 + 0 + 0.14 = 0.14 ] So, (P_X = {2:0.34, 3:0.33, 4:0.19, 5:0.14}). Marginal distribution of Y: [ P_Y(0) = 0 + 0.03 + 0 + 0 = 0.03 ] [ P_Y(1) = 0.34 + 0.30 + 0.16 + 0 = 0.80 ] [ P_Y(2) = 0 + 0 + 0.03 + 0.14 = 0.17 ] So, (P_Y = {0:0.03, 1:0.80, 2:0.17}). Conditional distribution of X given (Y=1): [ P(X=2|Y=1) = \frac{0.34}{0.80} = 0.425 ] [ P(X=3|Y=1) = \frac{0.30}{0.80} = 0.375 ] [ P(X=4|Y=1) = \frac{0.16}{0.80} = 0.20 ] [ P(X=5|Y=1) = \frac{0}{0.80} = 0 ] To test independence, check if (P(X=x, Y=y) = P_X(x) P_Y(y)) for all x, y. For example, (P(X=2,Y=1)=0.34) and (P_X(2)P_Y(1) = 0.34 \times 0.80 = 0.272 \neq 0.34). Hence, X and Y are not independent. --- For (Z = \begin{pmatrix} 2X_1 + X_2 - X_3 \ X_1 + X_2 + X_3 \end{pmatrix}), if (X \sim N_3(\mu, \Sigma)), then (Z) is also
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