Discrete Random Variables - pmf

The discrete random variable R takes the values in S = { -3, -1, 1, 3 } with probabilities respectively,

( 1 - theta )/4, ( 1 - 3theta) /4, ( 1 + 3theta)/4, (1+ theta)/4,

where theta is a real constant. Find the range of values for which this is a valid probability mass function.

i think the answer is -1/3 =< theta =< 1/3

BUT HOW DO PROVE IT!!!!!!!

help!!

Re: Discrete Random Variables - pmf

Quote:

Originally Posted by

**Matt1993** The discrete random variable R takes the values in S = { -3, -1, 1, 3 } with probabilities respectively,

( 1 - theta )/4, ( 1 - 3theta) /4, ( 1 + 3theta)/4, (1+ theta)/4,

where theta is a real constant. Find the range of values for which this is a valid probability mass function.

i think the answer is -1/3 =< theta =< 1/3

BUT HOW DO PROVE IT!!!!!!!

help!!

Hi Matt1993! :)

What is the reason you think that -1/3 =< theta =< 1/3?

That is likely the key to the proof...

Re: Discrete Random Variables - pmf

See all I did was let the pmfs equal zero and then I just tried values until I came up with that answer. That's the problem

Re: Discrete Random Variables - pmf

Quote:

Originally Posted by

**Matt1993** See all I did was let the pmfs equal zero and then I just tried values until I came up with that answer. That's the problem

That's close.

The axioms of probability (see wiki) require in particular 2 things from the probabilities:

- Each probability is at least 0: $\displaystyle p \ge 0$.

. - The sum of all probabilities is 1: $\displaystyle \sum p = 1$.

In your case that means:

$\displaystyle ( 1 - \theta )/4 \ge 0$

$\displaystyle ( 1 - 3\theta) /4 \ge 0$

$\displaystyle ( 1 + 3\theta)/4 \ge 0 $

$\displaystyle (1+ \theta)/4 \ge 0$

$\displaystyle ( 1 -\theta )/4 + ( 1 - 3\theta) /4 + ( 1 + 3\theta)/4 + (1+ \theta)/4 = 1$