Lecture 5a

Connections across Lectures 1 to 4

Andrew Pua

2024-10-07

Plan for these slides

1. How does everything you have seen so far mesh together?
2. Introducing the idea of random variable as a way to link the sample space to data
3. Ways to describe random variables
4. Your first special distribution
5. Connections to hypothesis tests you have learned so far

How do we link the sample space to the data?

• Sample spaces and probabilities are idealizations. In our context, an idealization is something that arises in the long-run or in the population.

• We are now going to study another idealization. You have already observed quantities in the real world that are subject to randomness.

• Examples include:

  • Measurements obtained from experiments
  • Measurements obtained from observational data
  • Statistics used for summaries
  • Statistics used to evaluate claims

Concept of a random variable

• Two students are selected at random (with replacement) from a college in which 60% of the students are male. Let \(X\) be the number of males in the sample.

• Define \(M\) to be a label for male, and \(F\) for female.

• Here we have \[\Omega=\{MM, MF, FM, FF\}.\]

• Here \(X\left(MM\right)=2\). What happens to the other elements of \(\Omega\)?

• Here \(X\) is what is called a random variable. It connects a particular outcome in a sample space with a value for \(X\).

Definition 1 (Wasserman (2004, p. 19) Definition 2.1) A random variable is a mapping \(X:\Omega\to \mathbb{R}\) that assigns a real number \(X(\omega)\) to each outcome \(\omega\).

• In effect, we have “hidden” the sample space in order to directly work with a suitable random variable of interest.

Concept of a distribution of a random variable

• A list of values of \(X\), together with the corresponding probabilities is called the distribution of \(X\).

• In effect, we “distribute” 1 to the possible values that a random variable \(X\) can take.

  • \(\mathbb{P}\left(X\in A\right)=\mathbb{P}\left(\{\omega\in\Omega: X(\omega)\in A\}\right)\) for a subset \(A\) of \(\mathbb{R}\)
  • \(\mathbb{P}\left(X=x\right)=\mathbb{P}\left(\{\omega\in\Omega: X(\omega)=x\}\right)\) for a particular value \(x\) of \(X\)

• For example, \[\mathbb{P}\left(X=2\right)= \mathbb{P}\left(\{MM\}\right)=0.6^2.\]

• Now finish constructing the distribution of \(X\), where \(X\) is the number of males in the sample.

• Given the distribution of \(X\), you can also find \(\mathbb{P}\left(X\leq 1\right)\). You can also generalize to \(\mathbb{P}\left(X\leq x\right)\) for any \(x\in\mathbb{R}\).

Cumulative distribution functions

Definition 2 (Wasserman (2004, p. 20) Definition 2.2) The cumulative distribution function or cdf is the function \(F_X:\mathbb{R}\to [0,1]\) defined by \[F_X\left(x\right) = \mathbb{P}\left(X\leq x\right).\]

Connections to previous lectures

A fair coin is tossed independently for ten times.

• Find the distribution of the \(X\), defined as the number of heads in those ten tosses.
• Change the coin from a fair one to the situation where the probability of heads is 0.25.
• Change the coin from a fair one to the situation where the probability of heads is \(p\) and instead of 10, consider \(n\) tosses.

Outcome \(\omega\)	Probability	\(X(\omega)\)
TTTTTTTTTT	\((0.5)^{10}\)	0
TTTTTTTTTH	\((0.5)^9 (0.5)\)	1
TTTTTTTTHT	\((0.5)^9 (0.5)\)	1
\(\vdots\)	\(\vdots\)	\(\vdots\)
HTTHHTHHH	\((0.5)^4 (0.5)^6\)	6
\(\vdots\)	\(\vdots\)	\(\vdots\)
HHHHHHHHHH	\((0.5)^{10}\)	10

• Eventually, you will obtain the distribution of \(X\), defined as the number of heads:

\[\mathbb{P}\left(X=k\right)=\begin{pmatrix}10 \\ k \end{pmatrix} (0.5)^k (0.5)^{10-k}, \ k=0,1,\ldots, 10\]

• What happens when you consider the other modifications?
• Congratulations! You have formed a special or named distribution called a binomial distribution.

• Experiments resulting in “success” or “failure” but not both
• You have to define what “success” means and ensure that “failure” is the complement.
• When \(n=1\), we have a Bernoulli distribution.
• Fix \(n\) as a positive integer and \(p\in (0,1)\). If there are \(n\) independent Bernoulli trials with each trial having the same probability of success \(p\), then the total number of successes \(X\) has a Binomial distribution with parameters \(n\) and \(p\). In short, \(X\sim Bin\left(n,p\right)\).

Connections to previous lectures

• 10 independent tosses of a coin

• Parallels to dogs example: 10 independent trials involving Harley choosing the correct cup

• Before we collect data:

  • We have 10 random variables \(X_1, X_2,\ldots, X_{10}\).
  • Here \(X_i\) takes on only two possible values 0 and 1.
  • Assume that \(\mathrm{P}\left(X_i=1\right)=\theta\) for all \(i\) and that each trial is independent of the next.

• We wanted to the test the null hypothesis that \(\theta=0.5\) against the alternative that \(\theta>0.5\).

  • To evaluate which hypothesis is supported by the data, we computed a \(p\)-value, which depends on the distribution of the sample proportion.
    
  • The sample proportion is the number of times Harley chose the correct cup is: \[ \overline{X}_{10} =\frac{X_1+X_2+\ldots+X_{10}}{10}.\]

• We used simulation to compute a \(p\)-value which is given by \[\mathbb{P}_{\theta=0.5}\left(\overline{X}_{10} \geq 0.9 \right)\] or \[\mathbb{P}_{\theta=0.5}\left(10\overline{X}_{10} \geq 9 \right)=\mathbb{P}_{\theta=0.5}\left(X_1+X_2+\ldots+X_{10} \geq 9 \right).\]

• What distribution springs to mind?

• Can you now compute (by hand and using R) the exact \(p\)-value for Harley’s case?