Tuesday, Jan 27

This is an outline of the topics we covered in class. These notes are not a substitute for your own note-taking. I highly recommend that you take your own notes during class. If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.

Continuous Distributions

Let X\in [0, 1] with f(x) = kx^{3/2}. Find k. 1 = \int_0^1 f(x)\ dx \amp = \int_0^1 kx^{3/2} dx = \frac{2kx^{5/2}}{5}\bigg|_0^1 = \frac{2k}{5} - 0 \Rightarrow \quad 1 \amp = \frac{2k}{5} \quad \Rightarrow \quad k = \frac{5}{2} Then, we can calculate probabilities, e.g.: \Pr\left(X \lt \frac{1}{2}\right) = \int_0^{1/2} \frac{5}{2}x^{3/2}\ dx = \frac{5}{2}\cdot \frac{2x^{5/2}}{5}\bigg|_0^{1/2} = \left(\frac{1}{2}\right)^{5/2} \approx 0.177.

Note: a pdf outputs probability densities, not probabilities. To get probabilities, we must integrate.

Let X be a continuous random variable with values in [a, b]. The cumulative distribution function (cdf) is: F(x) = \Pr(X \leq x). To calculate it: F(x) = \int_a^x f(t)\ dt.

Let X \in [0, 1], f(x) = \frac{5}{2}x^{3/2}. Then: F(x) = \int_0^x f(t)\ dt \amp = \int_0^x \frac{5}{2}t^{3/2}\ dt = \frac{5}{2}\cdot \frac{2t^{5/2}}{5}\bigg|_0^x F(x) \amp = x^{5/2} Then, we can calculate probabilities, e.g.: \Pr\left(X \lt \frac{1}{2}\right) \amp = F\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{5/2} \approx 0.177 \Pr(0.2 \leq X \leq 0.6) \amp = F(0.6) - F(0.2) = (0.6)^{5/2} - (0.2)^{5/2} \approx 0.261

Given f(x), we can find F(x) by calculating: F(x) = \int_a^x f(t)\ dt. Given F(x), we can find f(x) by calculating: f(x) = F'(x).

Suppose a machine needs repairs on average twice per month. Let T be the time until repair. This is a Poisson process (\lambda = 2 per month).

T is said to have the exponential distribution with parameter \lambda. We'll write T \sim \Exp(\lambda). The exponential density function is: f(t) = \lambda e^{-\lambda t}, \quad t \geq 0.

Let \lambda = 2, f(t) = 2e^{-2t}, t\geq 0. Then: F(t) \amp = \int_0^t f(x)\ dx = \int_0^t 2e^{-2x}\ dx = -e^{-2x}\bigg|_0^x \amp = -e^{-2t} - (-e^0) = 1 - e^{-2t}. So, e.g.: \Pr\left(T \leq \frac{3}{4}\right) \amp = F\left(\frac{3}{4}\right) = 1 - e^{-3/2} \approx 0.777 \Pr(T \gt 1) \amp = 1 - \Pr(T \leq 1) = 1 - F(1) \amp = 1 - (1 - e^{-2}) = e^{-2} \approx 0.135.

The distributions Bin, Geom, Poiss, and Exp are conceptually linked.

Relationship of common distributions Counting Events Time Until Discrete Time Bin Geom Continuous Time Poiss Exp

Joint Distributions

Let X take values x_1, x_2, \dotsc, x_n and Y take values y_1, y_2, \dotsc, y_m. The joint distribution of X and Y is the collection of all values \Pr(X = x_i, Y = y_j) for every i, j combination. The separarte distributions for X and Y are called marginal distributions.

Suppose X \in \{1, 2, 3\}, Y \in \{0, 1\}, with joint distribution below.

Example Joint Distribution X = 1 X = 2 X = 3 Y = 0 0.1 0.15 0.05 Y = 1 0.2 0.2 0.3

We find the marginal distribution for X by summing along the columns: \Pr(X = 1) \amp = 0.3 \Pr(X = 2) \amp = 0.35 \Pr(X = 3) \amp = 0.35 We find the marginal distribution for Y by summing along the rows: \Pr(Y = 0) \amp = 0.3 \Pr(Y = 1) \amp = 0.7

Random variables X, Y are independent if: \Pr(X = x_i, Y = y_j) = \Pr(X = x_i)\Pr(Y = y_j) for every i, j combination.

In the previous example: \Pr(X = 2, Y = 1) = 0.2 \neq (0.35)(0.7) = \Pr(X = 2)\Pr(Y = 1), so X, Y are not independent.

Let X, Y indicate heads on the first and second flip, respectively, of a fair coin. Then:

Joint Distribution for Indicator Random Variables X = 0 X = 1 Y = 0 1/4 1/4 Y = 1 1/4 1/4

Then the marginal distributions are: \Pr(X = 0) \amp = \frac{1}{2} \amp \Pr(Y = 0) \amp = \frac{1}{2} \Pr(X = 1) \amp = \frac{1}{2} \amp \Pr(Y = 1) \amp = \frac{1}{2} Then, for any a, b: \Pr(X = a, Y = b) = \frac{1}{4} = \frac{1}{2}\cdot \frac{1}{2} = \Pr(X = a)\Pr(Y = b). So X, Y are independent.