Question

Given 2 independent random vars X & Y. -2 -1 3 -1 0 2 and 0.2 0.3 0.5 0.2 0.4 0.4
Show distribution and calculate Expected value and variation of random variables 3X-2Y, X*x+y*y and X*Y.

Answer 1

Given 2 independent random vars X and Y. -2 -1 3 -1 0 2 and 0.2 0.3 0.5 0.2 0.4 0.4. Show distribution and calculate Expected value and variation of random variables 3X-2Y, X*x+y*y and X*Y.

Expected value:

X:

$$- 2 \cdot \left(\frac {1}{6}\right) - 1 \cdot \left(\frac {1}{3}\right) + 2 \cdot \left(\frac {1}{6}\right) + 3 \cdot \left(\frac {1}{6}\right) = - \frac {1}{3} + \frac {1}{2} = \frac {1}{6}$$

Y:

$$0. 2 \cdot \left(\frac {1}{3}\right) + 0. 4 \cdot \left(\frac {1}{3}\right) + 0. 5 \cdot \left(\frac {1}{6}\right) + 0. 3 \cdot \left(\frac {1}{6}\right) = \frac {1}{3}$$

Variation:

X:

$$X _ {m i n} = - 2, X _ {m a x} = 3$$

Y:

$$Y _ {m i n} = 0. 2, Y _ {m a x} = 0. 5$$

1. For $Z = 3X - 2Y$ :

a. Expected value: $\frac{3}{6} -\frac{2}{3} = -\frac{1}{6}$

b. Variation: $Z_{min} = (-2 \cdot 3 - 0.2 \cdot 2) = -6.4, Z_{max} = (3 \cdot 3 - 0.5 \cdot 2) = 8$

2. For $Z = X^2 + Y^2$ :

a. Expected value: $\frac{1}{36} +\frac{1}{9} = \frac{5}{36}$

b. Variation: $Z_{min} = ((-2)^{2} + (0.2)^{2}) = 4.04, Z_{max} = ((3)^{2} + (0.5)^{2}) = 9.25$

3. For $Z = XY$ :

a. Expected value: $\frac{1}{6} \cdot \frac{1}{3} = \frac{1}{18}$

b. Variation: $Z_{min} = 0.2 \cdot (-2) = -0.4, Z_{max} = 3 \cdot 0.5 = 1.5$