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I have a problem in my textbook:

Show that $\text{Cov}(X,Y)=(\text{Cov}(Y,X))^T$

My approach:

We have that $\text{Cov}(X,Y) = \mathbb{E}\biggl[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])^T\biggr]=\mathbb{E}\biggl[(Y-\mathbb{E}[Y])(X-\mathbb{E}[X])^T\biggr]^T=(\text{Cov}(Y,X))^T$

Would this be correct?

  • It looks correct as $A^TB = (B^TA)^T$. Btw, definition in wikipedia is given as $Cov(X,Y) = E[(X-E[X])(Y-E[Y])^T]$. You can also use $(A-B)^T = A^T - B^T$ to verify your solution. – Snowball May 11 '21 at 12:30
  • @Snowball I edited it. If possible, can you please elaborate what you mean by the $(A-B)^T$ part? Maybe extend this to an answer? –  May 11 '21 at 12:33
  • Your solution is correct since $E[X]^T = E[X^T]$. You can also expand $Cov(X,Y) = E[(X-\mu_X)(Y-\mu_Y)^T] = E[(X-\mu_X)(Y^T-\mu_Y^T)] = E[XY^T] - \mu_X\mu_Y^T = E[(YX^T)^T] - (\mu_Y\mu_X^T)^T = (E[YX^T] - \mu_Y\mu_X^T)^T = Cov(Y,X)^T$ – Snowball May 11 '21 at 12:45

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This problem can be solved using several approaches and your solution is also correct.

Method 1:

Since $AB^T = (BA^T)^T$, take $A = (X - E[X])$ and $B = (Y-E[Y])$ and use the fact $E[X]^T = E[X^T]$

Method 2:

Just expand everything

$Cov(X,Y) = E[(X-\mu_X)(Y-\mu_Y)^T] = E[(X-\mu_X)(Y^T-\mu_Y^T)] = E[XY^T] - \mu_X\mu_Y^T = E[(YX^T)^T] - (\mu_Y\mu_X^T)^T = (E[YX^T] - \mu_Y\mu_X^T)^T = Cov(Y,X)^T$.

Method 3:

Let $K = Cov(X,Y)$ and $K' = Cov(Y,X)$, then $K_{i,j} = E[(X_i - E[X_i])(Y_j - E[Y_j])] = K'_{j,i}$ and therefore, $K^T = K'$

Snowball
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