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A random variable (RV) is a random process with numerical outcomes
Random indicates that a single outcome of the process is uncertain.
Examples:
The sum of the dots that appear when two fair dice are rolled.
The number of patients that enter a hospital waiting room in any given hour interval.
The daily recorded high temperature in Corvallis.
The support of a random variable is the set of possible outcomes the RV can take on.
Discrete RVs
A random variable that has a countable set of possible outcomes.
finite OR
infinite with as many elements are there are whole numbers
Continuous RVs
A random variable that has an infinite, continuous set of possible outcomes in a given interval.
DISCRETE
\(x\) | 1 | 2 | 3 | 4 | 5 |
\(p(x)\) | 0.16 | 0.32 | 0.36 | 0.12 | 0.04 |
CONTINUOUS
Discrete RVs
Probability Mass Function (PMF), \(p(x)\)
\(p(x)\)\(= P(X = x)\)
The PMF may be represented as a
table
mathematical function
Properties:
\(0 \leq p(x) \leq 1\)
\(\sum \limits_{i=1}^n p(x_i) =1\)
Continuous RVs
Probability Density Function (PDF), \(f(x)\)
\(P(c \leq X \leq d) = \int \limits_{c}^d f(x) dx\)
Properties:
\(f(x) \geq 0\) for all values of \(x\) in the support
\(\int \limits_{-\infty}^{\infty} f(x) dx = 1\)
\(x\) | 1 | 2 | 3 | 4 | 5 |
\(p(x)\) | 0.16 | 0.32 | 0.36 | 0.12 | 0.04 |
\[p(2) = 0.32\]
\[p(4) = 0.12\]
Discrete RVs
\(F(x)\)\(=P(X \leq x)\)
\(F(x) = \sum \limits_{t\leq x}p(t)\)
Continuous RVs
\(F(x)\)\(=P(X \leq x)\)
\(F(x) = \int \limits_{-\infty}^x f(t) dt\)
Continuous CDF Properties:
\(P(a \leq X \leq b) = F(b) - F(a)\)
\(\frac{d}{dx}F(x) = f(x)\)
\(P(X \leq a) = P(X < a) = F(a)\)
\(x\) | 1 | 2 | 3 | 4 | 5 |
\(p(x)\) | 0.16 | 0.32 | 0.36 | 0.12 | 0.04 |
$$F(2)=P(X \leq 2)=p(1)+p(2)=0.16+0.32=0.48$$
\(F(x) =\)\(\int \limits_{0}^x 3t^2 dt\)\(=t^3 \bigg \vert _0^x\)\(=x^3\)
Discrete RVs
\[E(X) = \sum \limits_{i=1}^n x_i p(x_i)\]
Continuous RVs
\[E(X) = \int \limits_{-\infty}^{\infty} x f(x) dx\]
\(-\infty\) and \(\infty\) can be replaced with the bounds of the support of \(X\).
\(x\) | 1 | 2 | 3 | 4 | 5 |
\(p(x)\) | 0.16 | 0.32 | 0.36 | 0.12 | 0.04 |
\(E(X) = 1(0.16) + 2(0.32) + 3(0.36) + 4(0.12) + 5(0.04) = 2.56\)
We expect a randomly selected student at the community college to be enrolled in 2.56 courses.
\(E(X) =\)\(\int \limits _0^1 x (3x^2) dx\)\(= \frac{3}{4}x^4 \bigg \vert_0^1\)\(= \frac{3}{4}\)
We expect a randomly selected user of the reservation system occupies the meeting room for 3/4 of an hour (or 45 minutes).
Discrete RVs
\(Var(X) = \sum \limits_{i=1}^n(x_i-E(X))^2p(x_i)\) \(=E(X^2) - (E(X))^2\)
\(E(X^2) = \sum \limits_{i=1}^n x_i^2 p(x_i)\)
Continuous RVs
\(Var(X) = \int \limits_{-\infty}^{\infty}(x-E(X))^2f(x) dx\) \(=E(X^2) - (E(X))^2\)
\(E(X^2) = \int \limits_{-\infty}^{\infty} x^2 f(x) dx\)
\(-\infty\) and \(\infty\) can be replaced with the bounds of the support of \(X\).
\(SD(X) = \sqrt{Var(X)}\)
\(x\) | 1 | 2 | 3 | 4 | 5 |
\(p(x)\) | 0.16 | 0.32 | 0.36 | 0.12 | 0.04 |
\[Var(X) = E(X^2)-(E(X))^2\]
\[E(X^2) = 1^2(0.16) + 2^2(0.32) + 3^2(0.36) + 4^2(0.12) + 5^2(0.04) = 7.6\]
\[Var(X) = E(X^2)-(E(X))^2 = 7.6 - 2.56^2 = 1.0464\]
\[SD(X) = \sqrt{1.0464} = 1.022937\]
A typical deviation from the expected number of enrolled courses is approximately 1.02 courses.
\[Var(X) = E(X^2)-(E(X))^2\]
\(E(X^2)=\int \limits_0^1 x^2(3x^2)dx\)\(=\frac{3}{5}x^5 \bigg \vert_0^1\)\(= \frac{3}{5}\)
\[Var(X) = E(X^2)-(E(X))^2 = \frac{3}{5} - \bigg ( \frac{3}{4}\bigg)^2 = 0.0375\]
\[SD(X) = \sqrt{0.0375} = 0.1936\]
A typical deviation from the expected time spent in the reserved meeting room is approximately 0.19 hours (or a little more than 11 minutes).
Want to model the number of successful outcomes from \(n\) independent Bernoulli trials.
\(n=\) the number of independent Bernoulli trials
\(p=\) the probability of success on each independent trial
\[ p(x) = {{n}\choose{x}}p^x(1-p)^{n-x}\] for \(x\) in \(\{0, 1, 2, ..., n \}\)
where \({{n}\choose{x}} = \frac{n!}{x!(n-x)!}\)
\(E(X) = np\)
\(Var(X) = np(1-p)\)
Binomial Distribution
\(p(x) = P(X = x)\): dbinom(x, n, p)
\(F(x) = P(X \leq x)\): pbinom(q, n, p)