Question

In a reliability context a randomly selected electronic component will undergo an accelerated failure time test. Let X take the value 1 if the component lasts less than 50 hours and zero otherwise, and Y take the value 1 if the component lasts between 50 and 90 hours and zero otherwise. The probabilities that a randomly selected component will last less than 50 hours, between 50 and 90 hours, and more than 90 hours are 0.2, 0.5, and 0.3. Find the correlation of X and Y .

Answer 1

The probability distribution of X is,

P(X = 1) = 0.2

P(X = 0) = 1 - 0.2 = 0.8

The probability distribution of Y is,

P(Y = 1) = 0.5

P(Y = 0) = 1 - 0.5 = 0.5

E(X) = 0 * P(X = 0) + 1 * P(X = 1) = 0 * 0.8 + 1 * 0.2 = 0.2

E(Y) = 0 * P(Y = 0) + 1 * P(Y = 1) = 0 * 0.5 + 1 * 0.5 = 0.5

E(X²) = 0² * P(X = 0) + 1² * P(X = 1) = 0² * 0.8 + 1² * 0.2 = 0.2

E(Y²) = 0² * P(Y = 0) + 1² * P(Y = 1) = 0² * 0.5 + 1² * 0.5 = 0.5

Var(X) = E(X²) - E(X)²  = 0.2 - 0.2² = 0.16

Var(Y) = E(Y²) - E(Y)²  = 0.5 - 0.5² = 0.25

Now,  P(X = 0, Y = 0) = 0.3,   

P(X = 1, Y = 0) = 0.2

P(X = 0, Y = 1) = 0.5

E[XY] = 0 * 0 * P(X = 0, Y = 0) + 1 * 0 * P(X = 1, Y = 0) + 0 * 1 * P(X = 0, Y = 1) = 0

Cov(X, Y) = E[XY] - E[X] E[Y] = 0 - 0.2 * 0.5 = -0.1

Correlation of X and Y = Cov(X, Y) /

= -0.1 /

= -0.5