# Compute mean vector and covariance matrix

The random vector $$X = (X_1,X_2)^T$$ has mean $$\Bbb E[X] = \mu$$ and covariance matrix $$Cov(X) = \Sigma$$. It holds that

$$\Bbb E[B(X − b)] = (0,0)^T$$ and $$Cov(B(X − b)) = I_2$$

$$b= (1,2)^T$$ and $$B = \begin{pmatrix} -1 & 2 \\ 1 & -1 \\ \end{pmatrix}$$

Compute the mean vector and covariance matrix of $$X$$.

After plugging in the $$B$$ and $$b$$ into the expectation, I got that the mean vector is equal to $$b$$. I am not sure how to proceed with the covariance matrix.

Given :- $$E[B(X-b)] = (0,0)^T \implies E[BX] = [B][b] \implies E[X] = \mu_x = [B^{-1}][B][b] = [b] = (1,2)^T$$

Now we need to find $$:-$$ $$[Cov(X)] = E[XX^T] - (\mu_x)(\mu_x)^T ;$$ (The unknown is $$E[XX^T]$$ as $$\mu_x$$ is a known vector).

Consider $$X-b = Y$$ $$\implies$$ $$E[Y] = \mu_y = E[X-b] = E[X] - b = (0,0)^T$$

Given = $$[Cov(Y)] = E[YY^T] - (\mu_y)(\mu_y)^T$$ but as $$\mu_y = [0] \to [cov(Y)] = E[YY^T]$$

$$[cov(BY)] = [B][cov(Y)][B^T]$$ ....(You can verify this using the covariance formula)

But this is given to be $$[cov(BY)] = [I] \therefore$$ substituting the values $$\to [B][cov(Y)][B^T] = [I] \implies [cov(Y)] = [B^{-1}][B^T]^{-1} \implies E[YY^T] = [B^{-1}][B^T]^{-1}$$.

Now expanding $$E[YY^T]$$ by substituting back $$Y=X-b$$ we get $$\implies E[YY^T] = E[(X-b)(X-b)^T] = E[XX^T] - E[X][b^T] - [b]E[X^T] + [b][b^T]$$

$$\therefore E[YY^T] = E[(X-b)(X-b)^T] = E[XX^T] - \mu_x[b^T] - [b](\mu_x)^T + [b][b^T] = [B^{-1}][B^T]^{-1}$$

Everything is known except $$E[XX^T]$$. Get that quantity and substitute in the very first equation for $$[cov(X)]$$.