0

If our random variable only has six equal possible outcomes, will any probability distribution resulting in mapping to real numbers consist of only six real numbers each with probability $\frac{1}{6}$ and the rest of the real numbers with probability $0$?

For example:

We roll one dice. Let $Z$ be the number showing. Let $W=Z^3 + 4$ and $V=\sqrt{Z}$.

(a.) Compute $P(W=w)$ for every real number $w$. For this one, I plugged values $1:6$ for $Z$ and got that $P(W=w) = \frac{1}{6}$ when $w=5, 12, 31, 68, 129, 220$ and $P(W=w)=0$ otherwise.

(d.) Compute $P(VW=y)$ for every real number $y$. I also plugged values $1:6$ for $Z$ and got that $P(VW=y)=\frac{1}{6}$ when $y=5$, $12*\sqrt{2},31*\sqrt{3}, 136, 129*\sqrt{5},$ and $220*\sqrt{6}$ and $P(VW=y)=0$ otherwise.

There are other parts to the problem, but I'm confused because it seems like there will always be six real numbers each with a probability of $\frac{1}{6}$ because we only roll one dice.

Colbi
  • 815
SwabianOrtolan
  • 129
  • Please properly format the text using Mathjax latex – Mayank Deora Jun 07 '16 at 02:14
  • As long as the map is a bijection (onto and one-to-one), the support will consist of six distinct value and the distribution will have no bias. Let, say, $U= (2Z-5)^2$ and you will get quite a different result. – Graham Kemp Jun 07 '16 at 02:18

2 Answers2

1

Your answers are correct, but there won't always be six possibilities.

For example, let $Y=(Z-3)^2$. Then as $Z$ ranges from $1$ through $6$, $Y$ takes on the values $0$,$1$,$4$, and $9$ (only four values). This is because $Z=1$ and $Z=5$ both result in $Y=4$, and $Z=2$ and $Z=4$ both result in $Y=1$. So $P(Y=y)=\frac{1}{6}$ when $y=0,9$, $P(Y=y)=\frac{2}{6}=\frac{1}{3}$ when $y=1,4$, and $P(Y=y)=0$ otherwise.

kccu
  • 20,808
  • 1
  • 22
  • 41
  • 1
    This is what I thought: if the random variable maps to the same real number given two different inputs, then it makes intuitive sense that the probability would be 1/6+1/6. Thank you for the great example! I do not know how to use Mathjax latex, but I will look it up and try to use it next time. – SwabianOrtolan Jun 07 '16 at 02:29
0

There will not be more than $6$. It can be less than 6, for example, $P= \begin{cases} 0, & \text{ if } Z \text{ is even.} \\ 1, & \text{ if } Z \text{ is odd.}\end{cases}$

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149