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I have been having some trouble with "Indicator random variables". Here's an example:

It is known that $P \left( A \right) =0.7 $, $P \left( B\right) =0.6 $, and $P \left( A \cup B \right) =0.8 $ Find the Covariance between the Indicator Random Variables $1_{A}$ and $1_{A \cap B}$.

$P \left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P \left( A \cap B \right) \Rightarrow P \left( A \cap B \right)=0.5 $

For the rest... Any ideas would be appreciated.

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Does this need the homework tag? Actually, $0.7+0.6-0.7 \neq 0.5$, please check arithmetic. Also, do you know what covariance is? Edit your question with the definition and try to plug in the variables... – gt6989b Jun 27 '13 at 13:14
Its not homework. Covariance, yes I know what it is, the question is, how would you get the expected value from this probabilities, and as for the 0.5, I put the wrong value on the Union, now it should be fine. – Salieri Jun 27 '13 at 13:18
First write down the distributions for each variable. $1_A$ can take on the values 1 and 0 with probabilities 0.7 and 0.3 respectively. Then $$E[1_A]=1(0.7)+0(0.3)=0.7$$. Proceed similarly for the other variables and then compute the covariance in the usual way. – Wintermute Jun 27 '13 at 13:22
Thanks! When plugging the $E\left[ 1_{A} 1_{A\cap B} \right] $ into the definition of the covariance, how would you state it here? – Salieri Jun 27 '13 at 13:36
If I am right, $Cov \left( 1_{A}, 1_{A\cap B} \right) $ = $E\left( 1_{A} 1_{A\cap B} \right) $ - $E\left( 1_{A} \right) E \left( 1_{A\cap B} \right) $ – Salieri Jun 27 '13 at 13:44

1 Answer 1

Let $\Omega$ denote the set of possible outcomes. (Note that $A,B \subseteq \Omega$.)

We have $$ \mathbb{E}[1_A] = \int_\Omega 1_{\omega \in A} d\mathbb{P}(\omega) = \mathbb{P}[A]. $$ Note that $1_A^2 = 1_A$ therefore $\mathbb{E}[1_A^2] = \mathbb{E}[1_A] = \mathbb{P}[A]$.

Find the variance, do the same for $\mathbb{E}[1_{A\cap B}]$ and plug into the definition of covariance.

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If $E\left[ 1_{A}^{2} \right] $ = $E\left[ 1_{A} \right] $ shouldn't the variance be zero? – Salieri Jun 27 '13 at 13:31
@Cardonai For some random variable $X$, let $\mathbb{E}[X] = m = \mathbb{E}[X^2]$. Then $$Var(X) = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = m - m^2 = m(1-m),$$ which is not zero unless $m \in \{0,1\}$. – gt6989b Jun 27 '13 at 13:38
@Did thank you very much for the edit, I appreciate it. – gt6989b Jun 27 '13 at 13:47
Hey this is really helping thanks a lot @gt6989b – Salieri Jun 27 '13 at 13:48
@Cardonai Glad to be of help – gt6989b Jun 27 '13 at 13:59

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