It is known that P(x<10)=0.9 and P(|y|<1)=0.95 . Prove that…

It is known that $P(x\lt10)=0.9$ and $P(|y|\lt1)=0.95$. Prove that, whatever the dependence between $x$ and $y$, for $z=x+y$ the inequalities $P(z\lt11)\ge0.85$ and $P(z\lt9)\le0.95$ hold.

Step 1: The lower bound for $P(z\lt11)$

If $x\lt10$ and $|y|\lt1$, then $z=x+y\lt10+1=11$. Hence

\[ P(z\lt11)\ge P(x\lt10,\ |y|\lt1)\ge P(x\lt10)+P(|y|\lt1)-1=0.9+0.95-1=0.85. \]

\[ \boxed{P(z\lt11)\ge0.85.} \]

Step 2: The upper bound for $P(z\lt9)$

$P(x\ge10)=0.1$ and $P(y\gt-1)\ge P(|y|\lt1)=0.95$. If $x\ge10$ and $y\gt-1$, then $z\gt9$. So

\[ P(z\gt9)\ge P(x\ge10,\ y\gt-1)\ge0.1+0.95-1=0.05, \]

\[ \boxed{P(z\lt9)\le1-0.05=0.95.} \]

Step 3: The bounds can be reached

Split the sample space into three parts $E_1$, $E_2$, $E_3$ of probabilities $0.85$, $0.1$, $0.05$:

Here $P(x\lt10)=0.9$, $P(|y|\lt1)=0.95$, and $P(z\lt11)=0.85$ exactly: the first bound is reached.

For the second bound take four parts: probability $0.9$ with $x=0,y=0$ ($z=0$); $0.05$ with $x=10,y=0$ ($z=10$); $0.05$ with $x=10,y=-5$ ($z=5$). Then $P(x\lt10)=0.9$, $P(|y|\lt1)=0.95$ and $P(z\lt9)=0.9+0.05=0.95$ exactly.

So both bounds are sharp, and in general only the non-strict inequalities $P(z\lt11)\ge0.85$ and $P(z\lt9)\le0.95$ can be guaranteed.

Step 4: Meaning

Without knowing how $x$ and $y$ depend on each other, we can still bound the distribution of their sum using only the addition rule. This idea is used in reliability and risk analysis ("worst-case" bounds).