A random sample of $100$ variables is given. Each of them is independent and identically distributed with $N(0,1)$. What is the correlation between sum of $98$ variables and sum of $100$ variables?


Let $X_i \sim N(0,1)$ for $i \in \{1,\dots,100\}$ denote the random variables described in the question.

Next, define

$$ Y \equiv \sum_{i=1}^{98} X_i \;\;\;\;\; \text{and} \;\;\;\;\; Z \equiv \sum_{i=1}^{100} X_i $$

The correlation between $Y$ and $Z$ is given by

$$ \frac{\mathbb E [(Y-\mathbb E [Y])(Z-\mathbb E [Z])]}{ \sqrt{\mathbb E [(Y-\mathbb E [Y])^2] \mathbb E [(Z-\mathbb E [Z])^2]}} $$

Notice that $\mathbb E [Y] = \mathbb E [Z] = 0$, $\; \mathbb E[X_i X_j]=0$ for all $i\neq j, \;$ and $\mathbb E[X_i^2]=1$ for all $i\;$ which implies that

\begin{align} \mathbb E [(Y-\mathbb E [Y])(Z-\mathbb E [Z])] &= \sum_{i=1}^{98} \mathbb E[X_i^2]=98 \\[1.5ex] \mathbb E [(Y-\mathbb E [Y])^2] &= \sum_{i=1}^{98} \mathbb E[X_i^2]=98 \\[1.5ex] \mathbb E [(Z-\mathbb E [Z])^2] &= \sum_{i=1}^{100} \mathbb E[X_i^2]=100 \end{align}

and, therefore,

$$ \frac{\mathbb E [(Y-\mathbb E [Y])(Z-\mathbb E [Z])]}{ \sqrt{\mathbb E [(Y-\mathbb E [Y])^2] \mathbb E [(Z-\mathbb E [Z])^2]}} = \frac{98}{\sqrt{98*100}} $$

| cite | improve this answer | |
  • $\begingroup$ You found covariance. To find correlation, you need to divide by the product of standard deviations. $\endgroup$ – A.S. Nov 21 '15 at 19:14
  • $\begingroup$ @A.S. I corrected the mistake. $\endgroup$ – mzp Nov 21 '15 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.