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​​Discussion 10A

​​Expectation

​​​​$E[X] = \sum_x x \Pr[X = x] = \sum_x x \Pr_X[x]$​

​​​​$X$​ is a random variable and ​​$E[X]$​ is a scalar.

​​Physical Interpretation

​​If ​​$X$​ consists of masses on a number line, where the mass at position ​​$x$​ is ​​$\Pr[X=x]$​, then ​​$E[X]$​ is the center of mass of these masses. The value ​​$x \Pr[X=x]$​ can be thought of as the torque acting on the number line with a fulcrum at ​​$0$​. If the center of mass is at ​​$0$​ (​​$E[X]=0$​) then the net torque ​​$\sum_x x \Pr[X=x]$​ is ​​$0$​.

​​Variance

​​The variance of a random variable is a measure of its spread.

​​​​$\mathrm{Var}(X) = E[(X - E[X])^2] = E[X^2] - E[X]^2$​

​​Geometric Distribution

​​​​$X \sim \text{Geometric}(p)$​ when ​​$X$​ is the number of independent trials until a success, where each trial succeeds with probability ​​$p$​.

​​Proof of ​​$E[X] = \frac{1}{p}$​

​​With probability ​​$p$​ our trial succeeds, so ​​$X=1$​. Otherwise, we fail with probability ​​$1-p$​, meaning we need to start again, so ​​$X = 1 + X'$​, where ​​$X'$​ is a copy of ​​$X$​ (they have the same distribution). Then ​​$E[X] = (p) (1) + (1-p) (1 + E[X])$​. Rearranging gives us our result.

​​Poisson Distribution

​​If an event occurs with some average rate ​​$\lambda$​ instances per unit, then the number of times this event occurs over ​​$1$​ unit follows a Poisson distribution.

​​​​$X \sim \text{Poisson}(\lambda)$​

​​​​$\Pr[X=k] = \frac{\lambda^k}{k!} e^{-\lambda}$​

​​

​​The sum of two independent Poisson random variables ​​$X \sim \text{Poisson}(\lambda)$​ and ​​$Y \sim \text{Poisson}(\mu)$​ is ​​$X+Y \sim \text{Poisson}(\lambda + \mu)$​.

​​Ex. If an average of ​​$5.4$​ people walk through one door per hour and an average of ​​$3$​ people walk through the other door per hour, then the average number of people who walk through either door per hour is ​​$5.4 + 3$​.

​​Summary

​​

​​​​$X \sim \text{Bernoulli}(p)$​	​​​​$E[X] = p$​	​​​​$\mathrm{Var}(X) = p (1-p)$​
​​​​$X \sim \text{Geometric}(p)$​	​​​​$E[X] =\frac{1}{p}$​	​​​​$\mathrm{Var}(X) = \frac{1-p}{p^2}$​
​​​​$X \sim \text{Binomial}(n, p)$​	​​​​$E[X] = np$​	​​​​$\mathrm{Var}(X) = n p (1-p)$​
​​​​$X \sim \text{Poisson}(\lambda)$​	​​​​$E[X] = \lambda$​	​​​​$\mathrm{Var}(X) = \lambda$​

​​Tips and Tricks

​​If ​​$X \sim \text{Bernoulli}(p)$​ then ​​$X$​ has the same distribution as ​​$X^2 , X^3 , \cdots$​.

​​If ​​$Y \sim \text{Bernoulli}(q)$​ then ​​$X Y \sim \text{Bernoulli}(\Pr[X=1, Y=1])$​.

​​If they are independent then the parameter is ​​$\Pr[X=1, Y=1] = p q$​.

​​

​​If ​​$X = \sum_{i=1}^n X_i$​, then ​​$X^2 = \sum_{i, j} X_i X_j = \sum_{i=1}^n X_i^2 + \sum_{i \neq j} X_i X_j$​. If the ​​$X_i$​ have the same distribution, then ​​$E[X^2] = n E[X_i^2] + n (n-1) E[X_i X_j]$​.

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