Expectation and Variance

\(E[X|Y]\) is itself a random variable: a function of \(Y\) that gives the average of \(X\) for each value \(Y\) can take. Averaging it over the distribution of \(Y\) recovers the overall average of \(X\). For discrete variables,

\(E[E[X|Y]] = \sum_{y} E[X|Y = y]\,P(Y = y) = \sum_{y}\sum_{x} x\,P(X = x, Y = y) = \sum_{x} x\,P(X = x) = E[X]\).

This is the law of total expectation (the tower property). Option A is still random, not a fixed number, and B and D fail in general: if \(Y\) is independent of \(X\), then \(E[X|Y] = E[X]\), so the result is \(E[X]\) whatever \(E[Y]\) is.

Answer: C

Use \(E[Z] = 0\) and \(E[Z^{2}] = \operatorname{Var}(Z) = 1\).

• \(E[X] = aE[Z] + b = b\), so \(b = 1\).
• \(X - E[X] = aZ\), so \(E[(X - E[X])Z] = aE[Z^{2}] = a\), which gives \(a = -2\).
• Check: \(E[(X - E[X])^{2}] = a^{2}E[Z^{2}] = 4\) holds. This condition alone cannot fix the sign of \(a\); the second one does.

Option C has the right \(a\) but the wrong mean, while B and D fail the condition \(E[(X - E[X])Z] = -2\).

Answer: A

Standardising a normal variable gives \(Z \sim N(0, 1)\), so \(E[Z^{2}] = \operatorname{Var}(Z) = 1\). The fourth moment of a standard normal is \(E[Z^{4}] = 3\): its moment generating function \(e^{t^{2}/2}\) has \(t^{4}\) coefficient \(\frac{1}{8}\), and \(4! \cdot \frac{1}{8} = 3\). Therefore

\[\operatorname{Var}(Y) = E[Z^{4}] - \left(E[Z^{2}]\right)^{2} = 3 - 1 = 2.\]

Equivalently, \(Y = Z^{2}\) is chi-squared with 1 degree of freedom, whose variance is twice the degrees of freedom. Option C is \(E[Z^{4}]\) alone, not the variance.

Answer: B

Drawing with replacement leaves the bag unchanged, so the draws are independent and each is black with probability \(\frac{10}{15} = \frac{2}{3}\).

Write \(S_{100} = Y_{1} + \cdots + Y_{100}\), where \(Y_{k} = 1\) if the \(k\)-th draw is black and 0 otherwise. Each \(E[Y_{k}] = \frac{2}{3}\), so by linearity of expectation

\[ E[S_{100}] = 100 \times \frac{2}{3} = \frac{200}{3} \approx 66.67 \]

This is the mean \(np\) of a Binomial\((100, \frac{2}{3})\) variable. Rounded to one decimal place it is 66.7; using the white-ball probability \(\frac{1}{3}\) by mistake would give 33.3.

Answer: 66.7

Each throw is even with probability \(\frac{1}{2}\). Let \(E_{0}\) be the expected number of further throws at the start or just after an odd throw, and \(E_{1}\) the same just after an even throw.

• \(E_{1} = 1 + \frac{1}{2}(0) + \frac{1}{2}E_{0}\): an even throw finishes, an odd one starts over.
• \(E_{0} = 1 + \frac{1}{2}E_{1} + \frac{1}{2}E_{0}\), which gives \(E_{0} = 2 + E_{1}\).

Substituting, \(E_{0} = 3 + \frac{1}{2}E_{0}\), so \(E_{0} = 6\).

Option B (4) is the expected wait for two even throws that need not be consecutive (\(2 + 2\)); insisting that they come back to back costs two more throws on average.

Answer: C

For an exponential random variable with rate \(\lambda\):

\[ E(X) = \int_{0}^{\infty} x\lambda e^{-\lambda x}\,dx = \frac{1}{\lambda}, \qquad Var(X) = E(X^{2}) - E(X)^{2} = \frac{2}{\lambda^{2}} - \frac{1}{\lambda^{2}} = \frac{1}{\lambda^{2}} \]

The condition \(5E(X) = Var(X)\) becomes \(\frac{5}{\lambda} = \frac{1}{\lambda^{2}}\). Multiplying by \(\lambda^{2}\), which is allowed because \(\lambda > 0\), gives \(5\lambda = 1\), so \(\lambda = 0.2\).

Taking the mean as \(\lambda\) instead of \(1/\lambda\) is the usual slip; it would give \(\lambda = 5\).

Answer: 0.2