# Thread: Anyone Understand Random Variables?

1. ## Anyone Understand Random Variables?

For example, I am trying to solve the following problem:

If X1...Xn are iid random variables, then what values for "a" and "b" solve the following equation:

E[(X1+...+Xn)] = a*E[X1^2] + b*(E[X1])^2

While we're at it; my car insurance starts at 1000 dollars and reduces by 10% every year that I don't get into an accident. If there is only a 0.05 probability of me getting into an accident any year , does anyone know what the PMF for my total payments up to the year that I get into an accident will be ?

If anyone can help me, then I owe you one.

2. Originally Posted by anyt
For example, I am trying to solve the following problem:

If X1...Xn are iid random variables, then what values for "a" and "b" solve the following equation:

E[(X1+...+Xn)] = a*E[X1^2] + b*(E[X1])^2
Well, gigen that the Xi's are iid RV's:

E[(X1+...+Xn)] = n E[X1]

There is no unique solution for a and b such that:

n E[X1] = a*E[X1^2] + b*(E[X1])^2

but one that works is a=0, b=n/E[X1].

RonL

3. Originally Posted by anyt

While we're at it; my car insurance starts at 1000 dollars and reduces by 10% every year that I don't get into an accident. If there is only a 0.05 probability of me getting into an accident any year , does anyone know what the PMF for my total payments up to the year that I get into an accident will be ?
If the accident occurs in the $n$-th year $n \ge 1$, your payments will have been

$c(n)=1000 + 1000*0.9 + .. + 1000*0.9^{n-1}=1000~\frac{1-0.9^n}{1-0.9}$

and this occurs with probability:

$
p(c(n))=(1-0.05)^{n-1}\times 0.05
$

(that is the probability of no accidents in years $1$ to $n-1$, times the probability of an accident in year $n$) and all other payments have zero probability.

RonL

4. ## Thanks

Thanks for the help