Discrete Random Variable Probloem

  • #1

Homework Statement



Let X be a discrete random variable with probability mass function p given by:

a ...| -1 .| 0 ..| 1 ..| 2
-----+-----+-----+-----+---
p(a) | 1/4 | 1/8 | 1/8 | 1/2

and p(a) = 0 for all other a.

a.) Let random variable Y be defined by Y = X^2. Calculate the probability mass function of Y.

b.) Calculate the distribution functions for X and Y in a = 1, a = 3/4, a = pi - 3

Homework Equations



n/a

The Attempt at a Solution



a.) I know that if X = 2, Y = 4. And if X = 0, Y = 0, so
Py(4) = Px(2) = 1/2 and
Py(0) = Px(0) = 1/8

But what about -1 and 1? Does this mean that Py(1) = Px(-1) + Px(1)?

b.) Since we're only dealing with whole numbers, is it true that the probability distribution function for X and Y on a = 1, a = 3/4, a = pi - 3 is equal to PX(1) + PX(0) and PY(1) + PY(0) respectively?
 

Answers and Replies

  • #2
LCKurtz
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Yes, P(Y=1) = P(X^2=1) = P(X=+-1) = P(X=1) + P(X = -1)

For part b you aren't "dealing with whole numbers", whatever that means. The cumulative distribution functions for X and Y are defined for all real numbers. Your notation is confusing. I assume you are using Py and Px for the probability mass functions of X and Y. You need to give some notation for the cumulative distribution function (CDF). Let's call the CDF of X by the name F(x). We don't say "the function of X in a". You ask for F(a) which is P(X <= a). Once you have that straight you can probably tell whether your answers are correct.
 
  • #3
Thanks for your help! The terminology and the notation are really what get me. I'll have to keep working on it. Thanks again!
 

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