Question

The joint distribution of two variables X and Y are given below. Evaluate the following (i) Marginal distribution of X and Y (ii) E(XY) (iii) (iv) COV(X,Y),
(v) P(X + Y>1) and verify that X and Y are not independent.

Y
X 0 1
0 0.1 0.2
1 0.4 0.2
2 0.1 0

Answer 1

Answer on Question #53441 - Math -Statistics and Probability

The joint distribution of two variables $X$ and $Y$ is given below.

Table 1

Evaluate the following:

(i) Marginal distribution of X and Y;

(ii) $\operatorname {E}(\operatorname {XY})$

(iii) $\operatorname {COV}(\mathrm{X},\mathrm{Y})$

(iv) $\mathrm{P}(\mathrm{X} + \mathrm{Y} > 1)$

(v) verify that X and Y are not independent.

Solution

(i) X can take on values 0, 1 and 2.

Evaluate

$$P (X = 0) = P (X = 0, Y = 0) + P (X = 0, Y = 1) = 0. 1 + 0. 2 = 0. 3$$ $$P (X = 1) = P (X = 1, Y = 0) + P (X = 1, Y = 1) = 0. 4 + 0. 2 = 0. 6$$ $$P (X = 2) = P (X = 2, Y = 0) + P (X = 2, Y = 1) = 0. 1 + 0 = 0. 1$$

Y can take on values 0 and 1.

Evaluate

$$P (Y = 0) = P (X = 0, Y = 0) + P (X = 1, Y = 0) + P (X = 2, Y = 0) = 0. 1 + 0. 4 + 0. 1 = 0. 6$$ $$P (Y = 1) = P (X = 0, Y = 1) + P (X = 1, Y = 1) + P (X = 2, Y = 1) = 0. 2 + 0. 2 + 0 = 0. 4$$

So marginal distribution of X is

Table 2

So marginal distribution of Y is

Table 3

(ii) XY can take on values 0, 1, 2. Evaluate

$$\begin{array}{l} P(XY = 0) = P(X = 0, Y = 0) + P(X = 0, Y = 1) + P(X = 1, Y = 0) + P(X = 2, Y = 0) \\ = 0.1 + 0.2 + 0.4 + 0.1 = 0.8 \end{array}$$ $$P(XY = 1) = P(X = 1, Y = 1) = 0.2$$ $$P(XY = 2) = P(X = 2, Y = 1) = 0$$ $$E(XY) = \sum t \cdot P(XY = t) = 0 \cdot 0.8 + 1 \cdot 0.2 + 2 \cdot 0 = 0.2$$

(iii) Compute

$$E(X) = \sum t \cdot P(X = t) = 0 \cdot 0.3 + 1 \cdot 0.6 + 2 \cdot 0.1 = 0.8$$ $$E(Y) = \sum t \cdot P(Y = t) = 0 \cdot 0.6 + 1 \cdot 0.4 = 0.4$$ $$\mathrm{COV}(X,Y) = \mathrm{E}(XY) - \mathrm{EX} \cdot \mathrm{EY} = 0.2 - 0.8 \cdot 0.4 = 0.2 - 0.32 = -0.12$$

(iv) $X + Y$ can take on values 0,1,2,3

$$\begin{array}{l} P(X + Y > 1) = P(X + Y = 2) + P(X + Y = 3) = P(X = 2, Y = 0) + P(X = 1, Y = 1) + \\ + P(X = 2, Y = 1) = 0.1 + 0.2 + 0 = 0.3. \end{array}$$

(v) If X and Y are independent, then E(XY) = E(X) · E(Y), but in this problem that rule breaks, because E(XY) = 0.2 ≠ 0.32 = E(X) · E(Y). Thus, X and Y are not independent.

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