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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?

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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,

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

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  • $\begingroup$ 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

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