Q.8.3 - A First Course in Probability

Question

If $E\left[X\right]=75 E\left[Y\right]=75 Var\left(X\right)=10$

$Var\left(Y\right)=12 Cov(X,Y)=-3$

give an upper bound to 

(a) $P\left\{\right|X-Y|>15\};$

(b) $P\{X>Y+15\};$

(c) $P\{Y>X+15\}$;

Step-by-Step Solution

Answer

An upper bound to

(a). $P\left\{\right|X-Y|\ge 15\}\le \mathbf{.}\mathbf{1244}$

(b).$P\{X>Y+15\}\le \mathbf{.}\mathbf{1107}$

(c).$P\{Y>X+15\}\le \mathbf{.}\mathbf{1107}$

Part (a) Step 1: Given Information

Let $X$ and $Y$ be radnom variables such that

$E\left[X\right]=75,E\left[Y\right]=75,Var\left(X\right)=10$$Var\left(Y\right)=12,Cov(X,Y)=-3$

Part (a) Step 2: Explanation

Consider the random variable

$A=X-Y$

Since

${\mu }_A=E\left[A\right]=E\left(X\right)-E\left(Y\right)=0$

and

${\sigma }_A^2=Var\left(A\right)=Var\left(X\right)+Var\left(Y\right)-2Cov(X,Y)=28$

using Chebyshev's inequality we get

$P\left\{\right|X-Y|\ge 15\}=P\left\{\left|A-{\mu }_A\right|\ge 15\right\}\le \frac{{\sigma }_A^2}{{15}^2}=\frac{28}{{15}^2}=\mathbf{.}\mathbf{1244}$

Part (b) Step 1: Given Information

Let $X$ and $Y$ be random variables such that

$$\begin{gathered} E\left[X\right]=75,E\left[Y\right]=75,Var\left(X\right)=10 \\ Var\left(Y\right)=12, \\ Cov(X,Y)=-3 \end{gathered}$$

Part (b) Step 2: Explanation

Again, consider the random variable $A$. Now, using Proposition 5.l we get

$P\{X>Y+15\}=P\{A>15\}\le \frac{{\sigma }_A^2}{{\sigma }_A^2+{15}^2}=\frac{28}{28+{15}^2}=.\mathbf{1107}$

Part (c) Step 1: Given Information

Let $X$ and $Y$ be radnom variables such that

$E\left[X\right]=75,E\left[Y\right]=75,Var\left(X\right)=10$$Var\left(Y\right)=12,Cov(X,Y)=-3$

Part (c) Step 8: Explanation

If we denote as $B=Y-X$, then this random variable has mean

${\mu }_B=E\left[B\right]=E\left(Y\right)-E\left(X\right)=0$

and variance

${\sigma }_B^2=Var\left(B\right)=Var\left(Y\right)+Var\left(X\right)-2Cov(X,Y)=28$

Using One-sided Chebyshev inequality we get

$P\{Y>X+15\}=P\{B>15\}\le \frac{{\sigma }_B^2}{{\sigma }_B^2+{15}^2}=\frac{28}{28+{15}^2}=\mathbf{.}\mathbf{1107}.$