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 $Var[\textbf{X}]$.


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 $var(X_k)=k$.

  • $\begingroup$ Thanks for the suggestion. What about the covariances $Cov(X_{i},X_{j})$? $\endgroup$ – user1337 Mar 26 at 1:19
  • 2
    $\begingroup$ @user1337 Use the same representation of $X_i$ and $X_j$ and independence of summands. $\endgroup$ – NCh Mar 26 at 1:50

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