2023-10-25
\[ X = \left\{ \begin{array}{ll} 1 & \quad \mathrm{with\ probability}\ p \quad \\ 0 & \quad \mathrm{with\ probability}\ 1-p \quad \end{array} \right. \]
\[ p_X(x) = P(X=x) = p^x(1-p)^{1-x} \text{ for } x=0,1 \]
\[\text{E}(X) = p\]
\[\text{Var}(X) = pq = p(1-p)\]
Example 1
We roll a fair 6-sided die.
We get $1 if we roll a 5, and nothing otherwise.
Let \(X\) be how much money we get.
Find the mean and variance of \(X\).
Scenario: There are \(n\) independent trials, each resulting in a success or failure, with constant probability, \(p\), in each trial. We are counting the number of successes (or failures).
Shorthand: \(X \sim \text{Binomial}(n, p)\)
\[ X = \text{Number of successes of } n \text{ independent trials} \]
\[ p_X(x) = P(X=x) = {n \choose x}p^x(1-p)^{n-x} \text{ for } x=0,1,2, ..., n \]
\[\text{E}(X) = np\] \[\text{Var}(X) = npq = np(1-p)\]
Example 2
Suppose we roll a fair 6-sided die 50 times.
We get $1 every time we roll a 5, and nothing otherwise.
Let \(X\) be how much money we get on the 50 rolls.
Find the mean and variance of \(X\).
Scenario: There are repeated independent trials, each resulting in a success or failure, with constant probability of success for each trial. We are counting the number of trials until the first success.
Shorthand: \(X \sim \text{Geo}(p)\) or \(X \sim \text{Geometric}(p)\) or \(X \sim \text{G}(p)\)
| \(X =\) Number of trials needed for first success (count \(x\) includes the success) | \(X =\) Number of failures before first success (count \(x\) does not include the success) |
|---|---|
\(p _ X ( x ) = P(X=x) = (1-p)^{x-1}p\) for \(x=1,2, 3,...\) \(F_ X ( x ) = P(X\leq x) = 1-(1-p)^x\) for \(x=1,2, 3,...\) |
\(p _X (x)= P(X=x) = (1-p)^{x}p\) for \(x=0, 1,2,...\) \(F_X ( x ) = P(X\leq x) = 1-(1-p)^{x+1}\) for \(x=0, 1,2,...\) |
\(E(X)=\dfrac{1-p}{p}\) \(Var(X)= \dfrac{1-p}{p^2}\) |
\(E(X)=\dfrac{1-p}{p}\) \(Var(X) = \dfrac{1-p}{p^2}\) |
Example 3
We throw darts at a dartboard until we hit the bullseye. Assume throws are independent and the probability of hitting the bullseye is 0.01 for each throw.
What is the pmf for the number of throws needed to hit the bullseye?
What are the mean and variance for the number of throws needed to hit the bullseye?
Find the probability that our first bullseye:
is on the fourth try
is on one of the first four tries
is after the fifth try
is on one of the first fifty tries
is after the \(50^{th}\) try, given that it did not happen on the first 20 tries
Find the expected number of misses until we hit the bullseye.
Example 3
We throw darts at a dartboard until we hit the bullseye. Assume throws are independent and the probability of hitting the bullseye is 0.01 for each throw.
Example 3
We throw darts at a dartboard until we hit the bullseye. Assume throws are independent and the probability of hitting the bullseye is 0.01 for each throw.
Example 3
We throw darts at a dartboard until we hit the bullseye. Assume throws are independent and the probability of hitting the bullseye is 0.01 for each throw.
Find the probability that our first bullseye:
is on the fourth try
is on one of the first four tries
is after the fourth try
Example 3
We throw darts at a dartboard until we hit the bullseye. Assume throws are independent and the probability of hitting the bullseye is 0.01 for each throw.
Find the probability that our first bullseye:
is on one of the first fifty tries
is after the \(50^{th}\) try, given that it did not happen on the first 20 tries
If we know \(X\) is greater than some number (aka given \(X >j\)), then the probability of \(X > k+j\) is just the probability that \(X>k\).
\(P(X > k+j |X > j) = P(X > k)\) \[ P(X > k+j |X > j) = \dfrac{P(X>k+j \text{ and } X>j)}{P(X>j)} = \dfrac{P(X>k+j)}{P(X>j)} = \dfrac{(1-p)^{k+j}}{(1-p)^{j}} = (1-p)^{k} \]
\[ X = \text{Number of successes in } n \text{ draws} \]
\[ p_X(x) = P(X=x) = \dfrac{{M \choose x}{N-M \choose n-x}}{{N \choose n}} \] \[\text{ for } x \text{ integer-valued } \\ \max(0, n-(N-M)) \leq x \leq \min(n, M)\]
\[\text{E}(X) =\dfrac{nM}{N}\]
\[\text{Var}(X) = n \dfrac{M}{N} \bigg(1- \dfrac{M}{N} \bigg)\bigg(\dfrac{N-n}{N-1} \bigg)\]
Example 4
A wildlife biologist is using mark-recapture to research a wolf population. Suppose a specific study region is known to have 24 wolves, of which 11 have already been tagged. If 5 wolves are randomly captured, what is the probability that 3 of them have already been tagged?
Suppose a hypergeometric RV \(X\) has the following properties:
the population size \(N\) is really big,
the number of successes \(M\) in the population is relatively large,
and the number of items \(n\) selected is small.
Then, in this case, making \(n\) draws from the population doesn’t change the probability of success much, and the hypergeometric r.v. can be approximated by a binomial r.v.
Example 5
Suppose a specific study region is known to have 2400 wolves, of which 1100 have already been tagged.
If 50 wolves are randomly captured, what is the probability that 20 of them have already been tagged?
Approximate the probability in part (1) using the binomial distribution.
\[ X = \text{Outcome of interest, with } x=1, 2, ..., N \]
\[ p_X(x) = P(X=x) = \dfrac{1}{N} \text{ for } x=1, 2, 3, ..., N \]
\[\text{E}(X) =\dfrac{N+1}{2}\]
\[\text{Var}(X) = \dfrac{N^2 -1}{12}\]
Example 6
Examples of discrete uniform RVs
Chapter 14-16, 19-20 Slides