Q.6.1 - A First Course in Probability

Question

Verify Equation $\left(1.2\right)$.$P(a_1<X\le a_2,b_1<Y\le b_2)=F(a_2,b_2)+F(a_1,b_1)-F(a_1,b_2)-F(a_2,b_1)$

Step-by-Step Solution

Answer

Equation is proved.

$P(a_1<X\le a_2,b_1<Y\le b_2)=F(a_2,b_2)+F(a_1,b_1)-F(a_1,b_2)-F(a_2,b_1)$

Step 1 : Equation :

A declaration (represented by the symbol =) that the values of two mathematical expressions are equal.

Step 2 : Explanation :

Let us define the events,

$$\begin{gathered} A:\left\{X\le a_1\right\} \\ B:\left\{X\le a_2\right\} \\ C:\left\{Y\le b_1\right\} \\ D:\left\{Y\le b_2\right\} \end{gathered}$$

$$\begin{gathered} P\left[a_1<X<a_2\cap b_1\le Y\le b_2\right]=P\left[\left(B-A\right)\cap (D-C)\right] \\ =P\left[B\cap \left(D-C\right)-A\cap \left(D-C\right)\right]......\left(1\right) [By distributive law] \end{gathered}$$

We know that if $E\subset F\Rightarrow E\cap F=E$, then

$$\begin{gathered} P(F-E)=P(\overline{E}\cap F)=P\left(F\right)-P(E\cap F) \\ =P\left(F\right)-P\left(E\right).......\left(2\right) \end{gathered}$$

Obviously $A\subset B\Rightarrow \left[A\cap \left(D-C\right)\right]\subset \left[B\cap \left(D-C\right)\right]$

Hence by using $\left(2\right)$, we get from $\left(1\right)$

$$\begin{gathered} P\left[a_1<X\le a_2\cap b_1<Y\le b_2\right]=P\left[B\cap \left(D-C\right)\right]-P\left[A\cap \left(D-C\right)\right] \\ =P\left[\left(B\cap D\right)-\left(B\cap C\right)\right]-P\left[\left(A\cap D\right)-\left(A\cap C\right)\right] \\ =P\left(B\cap D\right)-\left(B\cap C\right)-P\left(A\cap D\right)+P\left(A\cap C\right)......\left(3\right) \end{gathered}$$

We have $P(B\cap D)=P\left[X\le a_2\cap Y\le b_2\right]=F\left(a_2,b_2\right)$

Similarly $$\begin{gathered} P(B\cap C)=F(a_2,b_1) \\ P(A\cap D)=F(a_1,b_2) \end{gathered}$$

Substitute these values in $\left(3\right)$ we get

$P(a_1<X\le a_2,b_1<Y\le b_2)=F(a_2,b_2)+F(a_1,b_1)-F(a_1,b_2)-F(a_2,b_1)$

Hence proved.