04. Random Variables
August 27th
Definition¶
A random variable (r.v.) is a probability space in which the outcomes are possible values of a variable (denoted by a capital letter: \(X,Y,Z…\))
Example
\(X=\) number of dots on a roll of a single die
\(\Omega=\{1,2,3,4,5,6\}\)
\(Y=\) number of dots on a roll two dice
\(\Omega=\{2,\dots,12\}\)
\(Z=\) number of times die is rolled until it lands on a 3
\(\Omega=\N_+\)
\(W=\) weight of newborn baby in kg.
\(\Omega=[1,4]\) note that this is an interval, not distinct values
\(X,Y,Z\) are discrete random values — countable (finite or subset of \(\N\))
\(W\) is a continuous r.v., the possible values form a continuous interval
The probability function \(f_x(k)\) of a discrete rv \(X\) gives the probability of \(X\) assuming each of its possible values, ie \(f_X(k)=P(X=k)\), meaning, the probability that \(X\) will indeed be \(k\).
Reiteration of previous example, now with NEW concept!
\(X\)= Roll of one die
\(f_X(2)=P(X=2)=\frac16\)
\(Y=\) Roll two dice
\(f_Y(2)=P(Y=2)=\frac1{36}\)
\(f_Y(3)=P(Y=3)=\frac2{36}\)
\(f_Y(4)=P(Y=4)=\frac3{36}\)
\(f_Y(5)=P(Y=5)=\frac4{36}\)
\(f_Y(6)=P(Y=6)=\frac5{36}\)
\(f_Y(7)=P(Y=7)=\frac6{36}\)
\(f_Y(8)=P(Y=8)=\frac5{36}\)
\(f_Y(9)=P(Y=9)=\frac4{36}\)
\(f_Y(10)=P(Y=10)=\frac3{36}\)
\(f_Y(11)=P(Y=11)=\frac2{36}\)
\(f_Y(12)=P(Y=12)=\frac1{36}\)
A probability function can be written as a list, or a table, or when possible, by a formula.
Going back to r.v. \(Z\), where \(Z\) is the number of rolls until we get a 3:
What is \(f_Z(k)\)?
\(\underbrace{\frac56\times\dots\times\frac56}_{k-1\text{ times}}\underbrace{\times\frac16}_{\text{rolling 3 on k-th try}}={(\frac56)}^{k-1}\cdot\frac16\)
Abstract
\(f_X(k)=\frac16\mid k\in\{1, 2, 3, 4, 5, 6\}\)
\(f_Y(k)=\) [list of values, see above example]
\(f_Z(k)={(\frac56)}^{k-1}\cdot\frac16\)
Since all outcomes \(X=k\) are disjoint and their union (over all possible values \(k\)) is \(\Omega\),
Exercise 4, Question 18
The number of flowers purchased by a customer in a certain flower store in a single purchase is a random variable \(X\), with the probability function:
Find the value of the constant \(k\).
(Do not confuse \(x\) with \(X\))
Rewrite formula as table:
| \(x\) | \(P_X(x)\) |
|---|---|
| 1 | \(\frac{k}{10}\cdot1^2=\frac{k}{10}\) |
| 2 | \(\frac{k}{10}\cdot2^2=\frac{4k}{10}\) |
| 3 | \(\frac{k}{10}\cdot3^2=\frac{9k}{10}\) |
| 4 | \(\frac{k}{10}\cdot4^2=\frac{16k}{10}\) |
| 5 | 0 |
| \(\vdots\) | 0 |
We know that the \(\sum_xP_X(x)=1\), that is, the sum of all values for \(P_X(x)\) must be one, so let’s add them up:
\(\frac{k}{10}+\frac{4k}{10}+\frac{9k}{10}+\frac{16k}{10}=1\)
Exercise 4, Question 16
If \(X\) is a random variable with the following probability function, find the value of \(a\):