Question

A biased coin has a probability of heads equal to \(1/3\) and a probability of tails equal to \(2/3\). A binary random variable \(X\) assumes a value \(1\) for heads and \(-1\) for tails. The variance of \(X\) is

• A. \(\frac{1}{3}\)
• B. \(\frac{5}{9}\)
• C. \(\frac{8}{9}\)
• D. \(1\)

Hint:

Use \(\text{Var}(X)=E[X^2]-(E[X])^2\), or treat \(X\) as a scaled Bernoulli variable.

Answer 1

The Correct Option is C

Step 1: Recall the definition of variance for a discrete random variable.
For a discrete random variable \(X\), the variance is
\[ \text{Var}(X) = E[X^2] - (E[X])^2 \]
We need \(E[X]\) and \(E[X^2]\) first.

Step 2: List the possible values and their probabilities.
\(X=1\) with probability \(P(\text{heads}) = \frac{1}{3}\).
\(X=-1\) with probability \(P(\text{tails}) = \frac{2}{3}\).

Step 3: Compute \(E[X]\).
\[ E[X] = (1)\left(\frac{1}{3}\right) + (-1)\left(\frac{2}{3}\right) = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3} \]

Step 4: Compute \(E[X^2]\).
Since \(X\) only takes the values \(1\) and \(-1\), \(X^2\) is always \(1\) regardless of the outcome.
\[ E[X^2] = (1)^2\left(\frac{1}{3}\right) + (-1)^2\left(\frac{2}{3}\right) = \frac{1}{3} + \frac{2}{3} = 1 \]

Step 5: Combine to get the variance.
\[ \text{Var}(X) = E[X^2] - (E[X])^2 = 1 - \left(-\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]

Step 6: Check why the other options are wrong.
Option (A) \(1/3\) is just \(P(\text{heads})\), not the variance. Option (B) \(5/9\) does not follow from any of the correct steps above; it could come from a sign or squaring slip. Option (D) \(1\) is \(E[X^2]\), the variance before subtracting \((E[X])^2\); forgetting that subtraction gives this wrong value.

Final Answer:
\[ \boxed{\frac{8}{9}} \]