If $X_i$,$Y_j$ are random variables, and $a_i$, $c_j$, $b$, $d$ $\in \mathbb{R}$, $i=1,2,...,n$; $j=1,2,...,m$, then show that
$$Cov \left( \sum_{i=1}^{n}a_i X_i+b,\sum_{j=1}^{m}c_j Y_j+d \right) =\sum_{i=1}^{n}\sum_{j=1}^{m}a_i c_j Cov(X_i,Y_j)$$

Is there a way to do this without induction? I started doing it that way, but the algebra is extremeeely messy.

Question

If $X_i$,$Y_j$ are random variables, and $a_i$, $c_j$, $b$, $d$ $\in \mathbb{R}$, $i=1,2,...,n$; $j=1,2,...,m$, then show that
$$Cov \left( \sum_{i=1}^{n}a_i X_i+b,\sum_{j=1}^{m}c_j Y_j+d \right) =\sum_{i=1}^{n}\sum_{j=1}^{m}a_i c_j Cov(X_i,Y_j)$$

Is there a way to do this without induction? I started doing it that way, but the algebra is extremeeely messy.

amistre64 · Answer

proof that the covariance is equal to that other stuff eh

amistre64 · Answer

im thinking an "easier" route might be to consider the algebra involved

amistre64 · Answer

define the formula for covariance,  input the given parts, and reduce it to the right side ....

kirbykirby · Answer

oh i see what you mean. I suppose that wold involve less algebra than induction, though still a bit messy I suppose. :)