# Given two random vectors, determine the dispersion matrix $Var[\textbf{X}]$.

Let $$\textbf{X} = (X_{1},X_{2},\ldots,X_{n})^{\prime}$$ be a vector of random variables, and let $$Y_{1} = X_{1}$$ and $$Y_{i} = X_{i}-X_{i-1}$$ where $$i = 2,3,\ldots,n$$. If $$Y_{i}$$ are mutually independent random variables, each with unity variance, find $$Var[\textbf{X}]$$.

MY ATTEMPT

To begin with, for $$i \geq 2$$, notice that \begin{align*} Var(Y_{i}) & = Cov(Y_{i},Y_{i}) = Cov(X_{i}-X_{i-1},X_{i}-X_{i-1})\\\\ & = Var(X_{i}) - 2Cov(X_{i},X_{i-1}) + Var(X_{i-1}) = 1\\\\ \end{align*}

On the other side, for $$i\neq j$$, we have \begin{align*} Cov(Y_{i},Y_{j}) & = Cov(X_{i}-X_{i-1},X_{j}-X_{j-1})\\\\ & = Cov(X_{i},X_{j}) - Cov(X_{i},X_{j-1}) - Cov(X_{i-1},X_{j}) + Cov(X_{i-1},X_{j-1}) = 0 \end{align*}

Unfortunately, I am not able to proceed from here. Any help is appreciated.

Verify that $$Y_1+Y_2+\cdots+Y_k=X_k$$ for each $$k$$. Since variance of a sum of independent random variables is the sum of the variances we get $$var(X_k)=k$$.
• Thanks for the suggestion. What about the covariances $Cov(X_{i},X_{j})$? – user1337 Mar 26 at 1:19
• @user1337 Use the same representation of $X_i$ and $X_j$ and independence of summands. – NCh Mar 26 at 1:50