How can the following covariance of two piece-wise defined random variables be calculated?

Let $K_1$ be a random variable defined piece-wise as follows:

$$K_1 = \begin{cases} a_1, & X_1 < H_1 \\ m_1 X_1+b_1, & H_1<X_1<H_2 \\ c_1, & X_1 > H_2 \end{cases}$$

$X_1$ is a random variable defined as:

$$X_1=L_1 (1+j_1 )$$

$L_1,H_1,H_2,a_1,m_1,b_1,c_1$ are arbitrary constants.

$$j_1\sim N(μ_1,σ_1^2)$$

Let $K_2$ be a random variable defined piece-wise as follows:

$$K_2 = \begin{cases} a_2, & X_2 < H_3 \\ m_2 X_2+b_2, & H_3 < X_2 < H_4 \\ c_2, & X_2 > H_4 \end{cases}$$

$X_2$ is a random variable defined as:

$$X_2=L_2 (1+j_2 )$$

$L_2,H_3,H_4,a_2,m_2,b_2,c_2$ are arbitrary constants.

$$j_2\sim N(μ_2,σ_2^2)$$

Problem. Calculate $\operatorname{Cov}(K_1,K_2)$.

I am able to calculate $E[K_1]$ and $E[K_2]$, but am unsure how to calculate $E[K_1K_2]$. Is there a way to calculate the above covariance without specification of a joint distribution? I'm assuming that $K_1$ and $K_2$ are not independent.

• Of course not. You have no hope of specifying $\operatorname{Cov}(K_1, K_2)$ without knowing the joint distribution of $(j_1, j_2)$. If you can assume that $j_1$ and $j_2$ are jointly normal, then you need to know only one extra input, the covariance $\operatorname{Cov}(j_1, j_2)$, to pin down the joint distribution of $(j_1, j_2)$ and hence the value of $\operatorname{Cov}(K_1, K_2)$. – Sangchul Lee Dec 25 '17 at 0:13
• Got it, thanks. – Alex Dec 25 '17 at 0:29