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Let $(X_1, X_2)$ be a randomly chosen pair out of $\{1,2, \ldots, 20\}$ (draw without repetition). Are both events $$E_1:=\{X_1 \geq 8\}$$ and $$E_2:=\{X_2 \geq 12\}$$ positive or negative correlated. Are they independent?

$$P(E_1\cap E_2) = \frac{9}{20}$$


$$P(E_1) \cdot P(E_2) = \frac{13}{20} \cdot \frac{9}{20} = \frac{117}{400}$$

on the basis of

$$P(E_1 \cap E_2) > P(E_1)\cdot P(E_2)$$

-> positiv correlated?

--- Add ---

We expect that both events $$(E_1\text{ and }E_2)$$ happen at the same time (the pair is chosen in one draw).

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$13/20+8/20\ne 13/50$; did you mean $(13/20)(8/20)$? In any case $P(E_2)=9/20$, and $P(E_1\cap E_2)\ne 8/20$. – Brian M. Scott Dec 7 '11 at 9:04
$$E_1 = {8,9,...20}$$ and $$E_2 = {12,13..20}$$ -> $$E_1 \cap E_2 = (12,13...20)$$ -> $$|E_1 \cap E_2| = 9 $$ or? – corium Dec 7 '11 at 11:58
up vote 2 down vote accepted

The notation $(X_1, X_2)$ means $X_1$ is the result of the first draw and $X_2$ the result of the second draw which occurs without replacement. Both $X_1$ and $X_2$ are uniformly distributed on $\{1, 2, \ldots, 20\}$, but they are not independent random variables. $$P(E_1) = P\{X_1 \geq 8\} = \frac{13}{20}, ~~ P(E_2) = P\{X_2 \geq 12\} = \frac{9}{20}.$$ But $E_1$ and $E_2$ are independent events if and only if $E_1^c$ and $E_2$ are independent events, in which case we would have $P(E_2 \mid E_1^c) = P(E_2)$. But clearly, $$P(E_2 \mid E_1^c) = \frac{9}{19} > \frac{9}{20} = P(E_2)$$ and so $E_1$ and $E_2$ are dependent events.

Since $P(E_1^c) = 1 - P(E_1)$, the law of total probability gives $$\begin{align*} P(E_2) &= P(E_2\mid E_1)P(E_1) + P(E_2\mid E_1^c)P(E_1^c)\\ &= P(E_2\mid E_1^c) + (P(E_2\mid E_1) -P(E_2\mid E_1^c))P(E_1) \end{align*}$$ which shows that $$\min\{P(E_2\mid E_1), P(E_2\mid E_1^c)\} \leq P(E_2) \leq \max\{P(E_2\mid E_1), P(E_2\mid E_1^c)\},$$ and since $P(E_2 \mid E_1^c) > P(E_2)$, we conclude that $$P(E_2 \mid E_1) < P(E_2) ~\text{and}~ P(E_2 \cap E_1) = P(E_2 \mid E_1)P(E_1) < P(E_2)P(E_1).$$ We can also calculate $$P(E_2 \mid E_1) = \frac{P(E_2) - P(E_2\mid E_1^c)P(E_1^c)}{P(E_1)} = \frac{\frac{9}{20}-\left(\frac{9}{19}\times\frac{7}{20}\right)}{\frac{13}{20}} =\frac{9 \times 12}{13 \times 19}$$ and so $$P(E_2 \cap E_1) = P(E_2 \mid E_1)P(E_1) = \frac{9 \times 12}{13 \times 19} \times \frac{13}{20} = \frac{9}{20}\times \frac{12}{19}$$ in contrast to the $P(E_2 \cap E_1) = \dfrac{9}{20}$ claimed by the OP.

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$E_1, E_2$ are not independent in the sense that if $E_1$ occurs, then the probability of $E_2$ occurs is less than that if $E_1$ does not occur. In other words, if the first number picked is greater than 8, it has chances to be also greater than 12, which limit the choices of second number. Therefore,$\frac{P(E_1\cap E_2)}{P(E_1)P(E_2)}=\frac{P(E_2|E_1)}{P(E_2)}\lt1$ and they are negatively correlated.

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In general, asking whether some variables are indepent OR positively OR negatively correlated is wrong, because non-correlation is a weaker property than independece (i.e. it could happen that they are not independent, but have zero correlation). But here we are speaking of events $E_1$ $E_2$, which can be assimilated to Bernoulli variables (taking the value 1 if event occurs). And in this case, it's true that independence coincides with zero correlation.

To see this, and to get the sign of the correlation, we compute the covariance:

$$Cov(e_1,e_2) = E[ \left(e_1 - E(e_1)\right) \left(e_2 - E(e_2)\right) ]=E[e_1 e_2] - E(e_1)E(e_2) = P(E_1 \cap E_2) - P(E_1)P(E_2)$$

which immediately shows that, $P(E_1 \cap E_2) > P(E_1) P(E_2)$ iff $E_1$ $E_2$ are positive ly correlated.

Updated: For the case in point, we have:

$ P(E_1) = \frac{13}{20}$

$ P(E_2) = \frac{9}{20}$

$ P(E_1 \cap E_2 ) = P (E_1 | E_2) P(E_2) = \frac{12}{19} \frac{9}{20}$

hence in our case $ P(E_1 \cap E_2 ) < P(E_1) P(E_2)$ which implies that the events are negatively correlated.

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We din't reach opposite conclusions. All I strived to show is that (in general) $E_1$ and $E_2$ are positively correlated $iff$ $P(E_1 \cap E_2) > P(E_1) P(E_2)$ (and viceversa: they are negatively correlated iff $P(E_1 \cap E_2) < P(E_1) P(E_2)$. Which was the issue raised by the OP in the last sentence (assuming the calculations of the concrete probabilities were ok). – leonbloy Dec 7 '11 at 16:59

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