Random Variable: A real-valued function defined on the sample space.
Range of Random Variable, R(X): All possible values X can take on.
Discrete Random Variable: A random variable with finite or countable R(X)
Two Types of Numerical Variables:
Probability Mass Function: P_X (i) = P(X = i), where i \in R(X)
Suppose we want to define X to be the number of heads we get after flipping a coin three times.
Our sample space would be:
In our sample space, heads can be gotten between 0—3 times. By putting the count above n, we can define the probability mass function in a table like so:
| i | P(X=i) |
|---|---|
| 0 | 1/8 |
| 1 | 3/8 |
| 2 | 3/8 |
| 3 | 1/8 |
| Total | 1 |
For example to find P(at most one heads), or P(X \le 1):
\begin{aligned} P(X \le 1) &= P(X = 0) + P(X = 1) \\ &= \frac{1}{8} + \frac{3}{8} \\ &= \frac{1}{2} \end{aligned}Q: Each of the following tables lists certain values of X and their probabilities P(x).
Determine whether each table represents a valid probability mass function.
| i | P(X=i) |
|---|---|
| 3 | 0.5 |
| 5 | 0.2 |
| 7 | 0.4 |
| i | P(X=i) |
|---|---|
| 0 | 0.3 |
| 1 | 0.9 |
| 2 | -0.2 |
| i | P(X=i) |
|---|---|
| 2 | 0.2 |
| 3 | 0.6 |
| 6 | 0.1 |
| 9 | 0.1 |
A:
Remember: A valid probability mass function will have its probabilities add up to 1.0, and each individual probability must be a valid probability (between 0 and 1).
Expectation of Random Variables; E(X) \lor \mu: Weighted average of a random variable’s outcomes.
\boxed{
E(X) = \sum_{i \in R(X)} i P(X=i) = \mu
} Q: Find the expectation of the random variable. Ex: Flip a coin three times. A: \begin{aligned}
E(X) &= \sum_{i \in \{ 0, 1, 2, 3 \}} i P(X=i)
&= 0P(X=0) + 1P(X=1) + 2P(X=2) + 3P(X=3)
&= 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8}
&= \frac{3}{2}
\end{aligned}Example: Random variable
i P(X=i) i P(X=i) 0 1/8 0 \times \frac{1}{8} 1 3/8 1 \times \frac{3}{8} 2 3/8 2 \times \frac{6}{8} 3 1/8 3 \times \frac{3}{8} Total 1 E(X) = \mu = \frac{3}{2}
Variance of Random Variables:
\boxed{
Var(X) = E(X^2) - \mu^2 = \sigma^2
} \\~\\
\small\textit{where } E(X^2) = \sum_{i \in R(X)} i^2 P(X=i)Example: Why?
\begin{aligned}
Var(X) &= E [ (X - \mu)^2 ] = \sum_{i \in R(X)} (i - \mu)^2 P(X-i) = \sigma^2 \\
&= E(X^2) - \mu^2 \quad \text{where } E(X^2) = \sum_{i \in R(X)} i^2 P(X=i)
\end{aligned}
Binomial Random Variable: X ~ B(n,p)
Assumptions:
\boxed{ \text{Binomial: } \begin{cases} R(x) &= \{ 0, 1, ..., n \} \\~\\ P(X=j) &= \binom{n}{j} p^j (1 - p)^{n-j} \\~\\ &\text{: Prob. of $j$ successes out of $n$ trials} \\~\\ E(X) &= np = \mu \\~\\ Var(X) &= \sqrt{np(1-p)} = \sigma \end{cases} }
Poisson Experiment: When we want to know the probability that a specific number of occurrences takes place within a given interval of time, length, or space (X).
Assumptions:
Poisson v.s. Binomial R.V.
- Total number of occurrences is fixed for Binomial, but Poisson has theoretically infinite range.
Q: Collect a sample of 10 people in a hospital and examine whether each of them has lung cancer.
A: X = number of patient with cancer. Number of trials is fixed. Binomial.
Q: Go to a hospital and see how many lung cancer patients come a day.
A: X = number of patient with cancer / day. No finite number of trials. Poisson.\lambda: \text{mean num. of occurrences during the given interval}
\boxed{
\text{Poisson: }
\begin{cases}
R(x) &= \{ 0, 1, 2, 3,4 , ... \} \\~\\
P(X=j) &= \frac{e^{-\lambda}\lambda^j}{j!} \\~\\
&\text{: Prob. of exactly $j$ occurrences in given interval} \\~\\
E(X) &= \lambda \\~\\
Var(X) &= \lambda \qquad (SD = \sqrt{\lambda})
\end{cases}
} Q: \lambda = 6, find P(X=4) A:
P(X=4) = \frac{e^{-6}6^4}{4!} = 0.1339 Q: \lambda = 6, find P(X \le 1) A:
P(X=0) = P(X=1) = 0.0025 + 0.0149 = 0.0174 Q: \lambda = 6, find P(X \ge 2) A:
P(X \ge 2) = 1 - P(X<2) = 0.9826 Given: A restaurant has on average 10 customers per day. Q: What is the probability they have exactly 5 customers on a given day? A: Let X: the # of customers a day ~ Pois(10) P(X=5) = \frac{e^{-10}10^5}{5!} = 0.0378 Q: What is the probability they have more than 3 customers on a given day? A:
P(X>3) = 1 - P(X \le 3) = 0.9896 Q: What is the expected number of customers for tomorrow? A: E(X) = 10 Q: What is the standard deviation of the number of customers? A:
We know Var(X) = 10Example: Poisson R.V.
X: The # of independent trials until the first success occurs with probability of success p.
\boxed{
\text{Geometric: }
\begin{cases}
R(x) &= \{ 1, 2, 3, 4 , ... \} \\~\\
P(X=j) &= (1-p)^{j-1} p, \qquad 0 < p < 1 \\~\\
E(X) &= 1/p \\~\\
Var(X) &= (1-p)/p^2
\end{cases}
} Roll a die until you get a 4. Let X: number of trials until 4 appears. Then,Example: Geometric R.V.
X: The number of objects with a specific trait out of a sample size n; where:
\boxed{
\text{Hypergeometric: }
\begin{cases}
R(x) &= \{ max(0,n+m-N), ..., min(m,n) \} \\~\\
P(X=j) &= \frac{\binom{m}{j} \binom{N-m}{n-j}}{\binom{N}{n}} \\~\\
E(X) &= \frac{nm}{N} \\~\\
Var(X) &= \frac{nm(N-n)(N-m)}{N^2(N-1)}
\end{cases}
} Q: There is an urn with 10 black balls and 6 white balls. 5 balls will be drawn in a row without replacement. Find the probability that two balls are white. A: Let X = Number of white balls R(X) = \{ 0, 1, ..., 5 \} \begin{align*}
P(X=2) &= \frac{\binom{m}{j} \binom{N-m}{n-j}}{\binom{N}{n}} \\
&= \frac{\binom{6}{2} \binom{16-6}{5-2}}{\binom{16}{5}} \\
&= 0.4127
\end{align*}Example: Hypergeometric R.V.