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Consider two random variables $X$ and $Y$ and their joint cumulative probability distribution function $F$. I'm attempting to show that $P\{a_1<X<a_2\mathrm{,}\, b_1<Y<b_2\} = F(a_2, b_2)+F(a_1,b_1)-F(a_2,b_1)-F(a_1,b_2)$ by simplifying the left hand side but I am not quite getting there.

My attempt $$P\{a_1<X<a_2\mathrm{,}\, b_1<Y<b_2\} = P(\{a_1<X<a_2\}\cap\{b_1<Y<b_2\}) = $$ $$P(\{X>a_1\} \cap \{X<a_2\} \cap \{Y>b_1\} \cap \{Y<b_2\}) = $$ $$1 - P((\{X>a_1\} \cap \{X<a_2\} \cap \{Y>b_1\} \cap \{Y<b_2\})^c) = $$ $$1 - P(\{X>a_1\}^c \cup \{X<a_2\}^c \cup \{Y>b_1\}^c \cup \{Y<b_2\}^c) = $$ $$1 - P(\{X\leq a_1\} \cup \{X\geq a_2\} \cup \{Y\leq b_1\} \cup \{Y\geq b_2\}) $$

So here I suppose I must apply the Inclusion–exclusion principle and becuase $\{X\leq a_1\}$, $\{X\geq a_2\}$ and $\{Y\leq b_1\}$, $\{Y\geq b_2\}$ are respectively mutually exclusive we have that the terms will only be probabilities defined on single sets and on the intersection of two non-mutually exclusive sets. Despite of this it still seems a bit too cumbersome and I am not sure how to simplify the probabilities of intersections.

$$1 - P(\{X\leq a_1\}) - P(\{X\geq a_2\}) - P(\{X\leq b_1\}) - P(\{X\geq b_2\}) $$ $$+ \,P(\{X\leq a_1\}\cap \{Y\leq b_1\}) + P(\{X\leq a_1\}\cap \{Y\geq b_2\}) $$ $$+ \,P(\{X\geq a_2\}\cap \{Y\leq b_1\}) + P(\{X\geq a_2\}\cap \{Y\geq b_2\}) $$

It just seems messy and I don't know how to continue. Am going in the right direction at least or am I completely off?

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\begin{equation} P(a_1<X<a_2,b_1<Y<b_2)=P(X<a_2,b_1<Y<b_2)-P(X<a_1,b_1<Y<b_2) \end{equation}

Also

\begin{equation} P(X<a_2,b_1<Y<b_2)=F(a_2,b_2)-F(a_2,b_1) \end{equation}

\begin{equation} P(X<a_1,b_1<Y<b_2)=F(a_1,b_2)-F(a_1,b_1) \end{equation}

And your result follows.

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  • $\begingroup$ Did you spot if I made any mistake in my attempt up to the point that I got to? $\endgroup$ – David Dec 20 '16 at 21:10
  • $\begingroup$ Not sure, seems your making it too complicated. $\endgroup$ – fesman Dec 20 '16 at 21:17

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