Random Variables and Expected Value

A random variable is just a number that depends on chance — like the sum of two dice or the number of heads in 3 coin flips. The expected value E(X) is the long-run average: if you repeated the experiment thousands of times, the average result would approach $E(X)$. It's a weighted average where each outcome is weighted by its probability. The variance measures how far outcomes tend to spread from that average. Think of it this way: expected value tells you what to "expect on average," and standard deviation tells you "how surprised you might be." For a linear transformation: $E(aX+b) = aE(X)+b$ and $Var(aX+b) = a^2 Var(X)$.

$$E(X)=\sum x\cdot p(x)$$

$$Var(X)=E(X^2)-[E(X)]^2, \quad \sigma=\sqrt{Var(X)}$$

Expected Value from a Table

Let $X=0,1,2,3$ with probabilities $0.10,0.30,0.40,0.20$. Find $E(X)$.

$$E(X)=0(0.10)+1(0.30)+2(0.40)+3(0.20)=1.70$$

Final answer: 1.70.

Variance of a Discrete Variable

For $X=1,2,3$ with probabilities $0.2,0.5,0.3$, find the variance.

$$E(X)=1(0.2)+2(0.5)+3(0.3)=2.1$$

$$E(X^2)=1(0.2)+4(0.5)+9(0.3)=4.9$$

$$Var(X)=4.9-(2.1)^2=0.49$$

Expected Profit

A project earns P80,000 with probability 0.35, P20,000 with probability 0.45, and loses P30,000 with probability 0.20. Find expected profit.

$$E=80000(0.35)+20000(0.45)-30000(0.20)=31,000$$

Final answer: P31,000.

Lottery Expected Value

A lottery ticket costs P50. The prize is P10,000 with probability 0.002 and P500 with probability 0.010. All other outcomes win nothing. Is the ticket a good buy?

Compute expected winnings (not including ticket cost):

$$E(\text{win})=10000(0.002)+500(0.010)+0(0.988)=20+5=25$$

Expected net = 25 − 50 = −P25. On average, you lose P25 per ticket. Final answer: Not a good buy — expected value is negative.

Standard Deviation of a Random Variable

A random variable X has the distribution: X = 0 (p = 0.3), X = 2 (p = 0.5), X = 4 (p = 0.2). Find the standard deviation.

$$E(X)=0(0.3)+2(0.5)+4(0.2)=0+1+0.8=1.8$$

$$E(X^2)=0(0.3)+4(0.5)+16(0.2)=0+2+3.2=5.2$$

$$Var(X)=5.2-1.8^2=5.2-3.24=1.96, \quad \sigma=\sqrt{1.96}=1.4$$

Final answer: $\sigma = 1.4$.

Linear Transformation of Expected Value

A random variable X has E(X) = 5 and Var(X) = 4. Find E(3X + 2) and Var(3X + 2).

Rules: $E(aX+b)=aE(X)+b$ and $Var(aX+b)=a^2Var(X)$. Adding a constant shifts the average but doesn't change the spread.

$$E(3X+2)=3(5)+2=17$$

$$Var(3X+2)=3^2(4)=36, \quad \sigma=6$$

Final answer: E = 17, Var = 36, σ = 6.

Verifying a Probability Distribution

Determine k so that the following is a valid probability distribution: P(X=1)=0.2, P(X=2)=k, P(X=3)=0.3, P(X=4)=0.1.

All probabilities must sum to 1 and each must be non-negative.

$$0.2+k+0.3+0.1=1 \implies k=0.4$$

Check: k = 0.4 ≥ 0 ✓. Final answer: k = 0.4.

Expected Sum of Two Dice

Two fair dice are rolled. Find the expected value and variance of the sum.

For one die: $E(\text{die})=3.5$ and $Var(\text{die})=\frac{35}{12}$. For independent dice:

$$E(\text{sum})=3.5+3.5=7$$

$$Var(\text{sum})=\frac{35}{12}+\frac{35}{12}=\frac{35}{6}\approx5.83$$

Final answer: E(sum) = 7, Var(sum) ≈ 5.83.