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 equation.

iv) Test the independence of variable s X and Y.

Let equation, where equation and

 


equation

Find the distribution of:



equation

v) Find equation.

10 Jan 2026
Answer :
Word Count : 550
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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