# Different solutions when using direct calculation and Chebychev's inequality

#### stasezer

##### New Member
Hi all,

I have a question that I can't find answer to:

I have 10 random variables X1, X2....X10 which are all independent and exponentially distributed with parameter=2

Xi~exp(2) for i between 1 and 10.

Now the argument says that the probability that the sum of all 10 X's is larger than 10 is less than 0.1.

P(sum of all X's > 10)<=0.1

if I calculate it with direct calculation I am doing it that way:

P(sum of all X's > 10) = P(X1+X2+...+X10 > 10) = P(X1+X1+...+X1 > 10) =
P(10X1 > 10) = P(X1 > 1) = 1 - P(X1 <= 1) =
1 - [integral from 0 to 1 of (2e^(-2*t)dt)]=
1 - 2((1/-2)*e^(-2*t)from 0 to 1) = 1 + (e^(-2) - e^(0)) = 1 + e^(-2) -1 =
e^(-2) = 0.13

and we get the answer that the argument is wrong( 0.1 < 0.13 ).

But, if we calculate the probability using Chebeyshev's inequality:

P(|X1+X2+...+X10 - E(X1+...+X10)| => 10 - E(X1+...+X10)) < V(X1+...+X10)/[10 - E(X1+...+X10)]^2

calculating E(X1+...+X10) = 10/2 = 5

calculating V(X1+...+X10) = 10/2^2 = 10/4 = 2.5

and now we can calculate
P(|X1+X2+...+X10 - E(X1+...+X10)| => 10 - E(X1+...+X10)) < 2.5/25 = 0.1

and we get that the argument is correct (0.1 <= 0.1)

I would really appreciate if someone could tell me where I am wrong.

Thanks !

#### katxt

##### Member
Unfortunately P(X1+X2+...+X10 > 10) is not the same as P(X1+X1+...+X1 > 10). In the second case you only need to get one number more than 1. In the first case you need to get several numbers more than 1 which is much harder.