Discrete random variables flashcards

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Discrete random variable
A quantity whose value is decided by chance and which can take only separate (countable) values, such as a count or a score.
Probability distribution of X
A list of every value x that X can take with its probability P(X = x), given as a table, a formula p(x) or a line diagram.
Probability function p(x)
A formula giving P(X = x) for each value x; p(x) = 0 for values X cannot take.
Cumulative distribution function F(x)
F(x₀) = P(X ≤ x₀): the running total of the probabilities up to and including x₀.
Mode of a discrete random variable
The value of x with the largest probability P(X = x).
Discrete uniform distribution
Every one of a finite set of values is equally likely; on 1, 2, …, n each has probability 1/n.
Conditions on a probability distribution
0 ≤ P(X = x) ≤ 1 for each x, and Σ P(X = x) = 1. (No — learn it)
E(X) for a discrete random variable
E(X) = μ = Σ x P(X = x). (Yes)
E(g(X))
E(g(X)) = Σ g(x) P(X = x); for example E(X²) = Σ x² P(X = x). (Yes)
Var(X)
Var(X) = E(X²) − [E(X)]² = Σ (x − μ)² P(X = x); standard deviation σ = √Var(X). (Yes)
E(aX + b) and Var(aX + b)
E(aX + b) = aE(X) + b; Var(aX + b) = a²Var(X). (No — learn it)
P(X = x) from F, whole-number values
P(X = x) = F(x) − F(x − 1). (No — derive it)
Mean and variance of the discrete uniform distribution on 1, …, n
E(X) = (n + 1)/2, Var(X) = (n² − 1)/12. (No — learn it)
Vertical line diagram of a distribution
A vertical line at each value x with height P(X = x); the heights add to 1 and the tallest line is the mode.
Graph of F(x) for a discrete random variable
A step graph: flat between values, jumping up at each value by P(X = x), from 0 to 1.
Line diagram of a discrete uniform distribution
All lines have the same height 1/n; the mean is in the middle.
Line diagram of Y = 2X + 1 compared with X
Same heights; every value doubled then moved up 1, so gaps and the standard deviation double and the mean becomes 2μ + 1.
Why can E(X) be a value X never takes?
It is a long-run average, the balance point of the distribution, e.g. 3.5 for a fair die.
Why does adding b not change Var(X)?
Every value moves by the same amount, so the distances from the mean, and hence the spread, are unchanged.
Why is a squared in Var(aX + b)?
Multiplying by a multiplies every distance from the mean by a, so every squared distance by a².
Is E(X²) equal to [E(X)]²?
No. Their difference is Var(X), which is positive unless X is constant.
How do you find two unknown probabilities in a table?
Use Σ P = 1 and one more condition, such as a given E(X); solve simultaneously and check each value is between 0 and 1.
Mean of a uniform distribution on 5, 6, …, 14
Write X = U + 4 with U uniform on 1, …, 10: E(X) = 5.5 + 4 = 9.5, Var(X) = 99/12 = 8.25.
P(X = x) = kx for x = 1, 2, 3, 4. Find k.
k(1 + 2 + 3 + 4) = 10k = 1, so k = 1/10.
X = 0, 1, 2 with probabilities 0.5, 0.3, 0.2. Find E(X) and Var(X).
E(X) = 0.3 + 0.4 = 0.7; E(X²) = 0.3 + 0.8 = 1.1; Var(X) = 1.1 − 0.49 = 0.61.
F(2) = 0.45 and F(3) = 0.8. Find P(X = 3).
F(3) − F(2) = 0.8 − 0.45 = 0.35.
E(X) = 5, Var(X) = 2. Find E(3X − 1) and Var(3X − 1).
E = 3 × 5 − 1 = 14; Var = 3² × 2 = 18.
E(X) = 3 and Var(X) = 4. Find E(X²).
E(X²) = Var(X) + [E(X)]² = 4 + 9 = 13.
Mean and variance of the score on a fair six-sided die?
Uniform, n = 6: mean 7/2 = 3.5, variance (36 − 1)/12 = 35/12.
X = 1 or 2, each with probability 1/2. Find E(1/X).
E(1/X) = (1)(1/2) + (1/2)(1/2) = 3/4, which is not 1/E(X) = 2/3.
After finding k in a probability function
Add all the probabilities and check the total is exactly 1 before going on.
P(X < 4) for a whole-number X, using F
P(X < 4) = P(X ≤ 3) = F(3), not F(4).
Var(5 − 2X) when Var(X) = 3
(−2)² × 3 = 12. The 5 disappears and the negative sign is squared away.
Variance came out negative
A slip: usually E(X)² was not subtracted correctly or a probability is wrong. A variance is never negative.
Using (n + 1)/2 and (n² − 1)/12
Only for values 1, 2, …, n. Shift or scale first if the values start elsewhere.

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