91Ó°ÊÓ

Covariance is a measure of the joint variability of two random variables in probability theory and statistics. The covariance is positive if the bigger values of one variable largely correlate to the greater values of the other variable, and vice versa for the lesser values (that is, the variables tend to behave similarly).

(a):

Proposition: The following holds

\({\mathop{\rm Cov}\nolimits} (X,Y) = E(XY) - E(X) \cdot E(Y).\)

The joint pmf of random variable \({X_1}\)and \({X_2}\)is given in the mentioned exercise. The marginal pmf of \({X_1}\)is

The Expected Value (mean value) of a discrete random variable X with set of possible values S and pmf of P(X) is

\(E(X) = {\mu _X} = \sum\limits_{x \in S} x \cdot p(x).\)

Therefore, the expected value is

\(\begin{aligned}E\left( {{X_1}} \right) &= 0 \cdot {p_{{X_1}}}(0) + 1 \cdot {p_{{X_1}}}(1) + 2 \cdot {p_{{X_1}}}(2) + 3 \cdot {p_{{X_1}}}(3) + 4 \cdot {p_{{X_1}}}(4)\\ &= 0 \cdot 0.19 + 1 \cdot 0.3 + 2 \cdot 0.25 + 3 \cdot 0.14 + 4 \cdot 0.12\\ &= 0 + 0.3 + 0.5 + 0.42 + 0.48\\ &= 1.7\end{aligned}\)

The marginal pmf of \({X_2}\)is

The expected value is

\(\begin{array}{c}E\left( {{X_2}} \right) = 0 \cdot 0.19 + 1 \cdot 0.3 + 2 \cdot 0.28 + 3 \cdot 0.23\\ = 1.55\end{array}\)

The following is expected values of random variable \({X_1}{X_2}\)

\(\begin{aligned}E\left( {{X_1}{X_2}} \right) &= \sum\limits_{{x_1}} {\sum\limits_{{x_2}} {{x_1}} } {x_2}p\left( {{x_1},{x_2}} \right)\\ &= 0 \cdot 0 \cdot 0.08 + 0 \cdot 1 \cdot 0.07 + \ldots + 4 \cdot 2 \cdot 0.05 + 4 \cdot 3 \cdot 0.06\\ &= 3.33\end{aligned}\)

The covariance can be computed as

\(\begin{aligned}{\mathop{\rm Cov}\nolimits} \left( {{X_1},{X_2}} \right) &= E\left( {{X_1}{X_2}} \right) - E\left( {{X_1}} \right)E\left( {{X_2}} \right)\\ &= 3.33 - 2.635\\ &= 0.695.\end{aligned}\)

(b):

Two random variables X and Y are independent if and only if

1.\(p(x,y) = {p_X}(x) \cdot {p_Y}(y)\),

for every (x, y) and when X and Y discrete rv's,

2.\(f(x,y) = {f_X}(x) \cdot {f_Y}(y)\),

for every (x, y) and when X and Y continuous rv's,

Otherwise, they are dependent.

By calculating probabilities

\(\begin{array}{l}P\left( {{X_1} = 4} \right) = {p_{{X_1}}}(4) = 0.12,\\P\left( {{X_2} = 0} \right) = {p_{{X_2}}}(0) = 0.19,\end{array}\)

and probability

\(P\left( {{X_1} = 4,{X_2} = 0} \right) = p(4,0) = 0\)

Notice that

\(p(4,0) = 0 \ne 0.0228 = 0.12 \cdot 0.19 = {p_{{X_1}}}(4) \cdot {p_{{X_2}}}(0),\)

from which we can conclude that the random variable is dependent.

Because the random variables are dependent, equality

\(V\left( {{X_1} + {X_2}} \right) = V\left( {{X_1}} \right) + V\left( {{X_2}} \right)\)

does not stand! Instead, use the following equality to compute the variance

\(V\left( {{X_1} + {X_2}} \right) = V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + 2{\mathop{\rm Cov}\nolimits} \left( {{X_1},{X_2}} \right).\)

First compute variances as follows

\(\begin{aligned}V\left( {{X_1}} \right) &= E\left( {X_1^2} \right) - \left( {E\left( {X_1^2} \right)} \right)\\ &= {0^2} \cdot 0.19 + {1^2} \cdot 0.3 + {2^2} \cdot 0.25 + {3^2} \cdot 0.14 + {4^2} \cdot 0.12 - {1.7^2}\\ &= 1.59\end{aligned}\)

and for random variable \({X_2}\)

\(\begin{aligned}V\left( {{X_2}} \right) &= E\left( {X_2^2} \right) - \left( {E\left( {X_2^2} \right)} \right)\\ &= {0^2} \cdot 0.19 + {1^2} \cdot 0.3 + {2^2} \cdot 0.28 + {3^2} \cdot 0.23 - {1.55^2}\\ &= 1.0875\end{aligned}\)

Finally, the following holds

\(\begin{aligned}V\left( {{X_1} + {X_2}} \right) &= V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + 2{\mathop{\rm Cov}\nolimits} \left( {{X_1},{X_2}} \right)\\ &= 1.59 + 1.0875 + 2 \cdot 0.695\\ &= 4.0675\end{aligned}\)