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Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $\in \Omega$ (the sample space):

How can I prove $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ using the equation: $E[\max(X,Y)] = E[X] + E[Y] - E[\min(X,Y)],\; A, B \in \Omega$ (my sample space)

I believe to prove this $E[\max(X,Y)] = E[X] + E[Y] - E[\min(X,Y)]$ I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?

And to use this equation to prove $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ I will have to use the idea of $A \subset B, \; B \subset A$ and $A = B,$ right?

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    $\begingroup$ you should porably explain your notations. $\endgroup$ – J.F Jan 18 at 20:33
  • $\begingroup$ @G.F Iedited the question. Thanks. $\endgroup$ – Laura Jan 18 at 20:36
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    $\begingroup$ I think you mean to write $$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$ but you have it miswritten in your problem $\endgroup$ – WaveX Jan 18 at 20:37
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    $\begingroup$ It helps to note that $A\cup B$ can be written as a disjoint union as $A\cup B = (A\setminus B)\cup (A\cap B)\cup (B\setminus A)$ and similarly that $A = (A\setminus B)\cup (A\cap B)$, etc... $\endgroup$ – JMoravitz Jan 18 at 20:37
  • $\begingroup$ @JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof? $\endgroup$ – Laura Jan 18 at 20:41
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If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $\max\{X,Y\}$ is the indicator variable of $A\cup B$, and $\min\{A,B\}$ of $A\cap B$; thus, $E(\max\{X,Y\})=P(A\cup B)$ and $E(\min\{X,Y\})=P(A\cap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(\max\{X,Y\})=E(X)+E(Y)-E(\min\{X,Y\})$ can be rewritten as $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.

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  • $\begingroup$ thanks @W-t-P! But how can I understand this concept of Indicator variable? $\endgroup$ – Laura Jan 19 at 1:20
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    $\begingroup$ @Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$ $\endgroup$ – WaveX Jan 19 at 3:41
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If you need to prove that $$ E[\max(X,Y)] = E[X] + E[Y] - E[\min(X,Y)], $$ simply note that the relation is valid even before you take expectations: $$ \max(X,Y) = X + Y - \min(X,Y), $$ which is the same as $$ \ X + Y = \min(X,Y) + \max(X,Y), $$ which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).

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