Discrete Random Variables: \(Y\) and function \(\mathrm{f}\) — EJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
- \(X\) is a random variable such that \(\mathrm{E}(X) = 4\) and \(\mathrm{E}\!\left[(X - 3)^2\right] = 40\). Find the value of \(\mathrm{Var}(X)\).
- Let \(Y\) be a discrete random variable. Let \(\mathrm{f}\) be a function which is defined for all values that \(Y\) can take. Then \(\mathrm{f}(Y)\) is also a random variable, and its expectation is given by \[ \mathrm{E}\!\left[\mathrm{f}(Y)\right] = \sum_{y}\!\left[\mathrm{f}(y) \times \mathrm{P}(Y = y)\right]. \] Suppose now that \(Y \sim \mathrm{B}\!\left(3, \tfrac{1}{3}\right)\) and \(\mathrm{f}(t) = \dfrac{1}{t + 1}\). Find the value of \(\mathrm{E}\!\left[\mathrm{f}(Y)\right]\).
- \(W\) is a random variable with the following probability distribution:It is given that \(\mathrm{E}(W) = -0.2\) and \(\mathrm{E}(W^3) = -3.2\). Find the values of \(p\), \(q\) and \(r\).
\(w\) \(-3\) \(-1\) \(1\) \(2\) \(\mathrm{P}(W = w)\) \(p\) \(q\) \(0.15\) \(r\)
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(a) Use \(\mathrm{Var}(X - 3) = \mathrm{Var}(X)\) (variance is translation-invariant): \[\begin{aligned} \mathrm{E}(X - 3) &= \mathrm{E}(X) - 3\\ &= 4 - 3 = 1,\\ \mathrm{Var}(X - 3) &= \mathrm{E}\!\left[(X-3)^2\right] - \left[\mathrm{E}(X-3)\right]^2\\ &= 40 - 1^2 = 39,\\ \mathrm{Var}(X) &= \mathrm{Var}(X - 3) = 39. \end{aligned}\]
Alternative method
Expand \((X - 3)^2 = X^2 - 6X + 9\): \[\begin{aligned} \mathrm{E}\!\left[(X-3)^2\right] &= \mathrm{E}(X^2) - 6\,\mathrm{E}(X) + 9 = 40,\\ \mathrm{E}(X^2) &= 40 + 6(4) - 9\\ &= 55,\\ \mathrm{Var}(X) &= \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\\ &= 55 - 16 = 39. \end{aligned}\]
\(\mathrm{Var}(X) = 39\).
(b) \(Y \sim \mathrm{B}\!\left(3, \tfrac{1}{3}\right)\), so \(Y\) takes values \(0, 1, 2, 3\). With \(\mathrm{f}(t) = \dfrac{1}{t+1}\):
| \(y\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(\mathrm{f}(y)\) | \(1\) | \(\tfrac{1}{2}\) | \(\tfrac{1}{3}\) | \(\tfrac{1}{4}\) |
| \(\mathrm{P}(Y=y)\) | \(\left(\tfrac{2}{3}\right)^3\) | \(\binom{3}{1}\!\left(\tfrac{2}{3}\right)^2\!\left(\tfrac{1}{3}\right)\) | \(\binom{3}{2}\!\left(\tfrac{2}{3}\right)\!\left(\tfrac{1}{3}\right)^2\) | \(\left(\tfrac{1}{3}\right)^3\) |
| \(\mathrm{f}(y)\,\mathrm{P}(Y=y)\) | \(\tfrac{8}{27}\) | \(\tfrac{2}{9}\) | \(\tfrac{2}{27}\) | \(\tfrac{1}{108}\) |
\(\mathrm{E}\!\left[\mathrm{f}(Y)\right] = \dfrac{65}{108} \approx 0.602\).
(c) From the three conditions: \[\begin{aligned} p + q + 0.15 + r = 1 &\quad\Longrightarrow\quad p + q + r = 0.85, \quad(1)\\ -3p - q + 0.15 + 2r = -0.2 &\quad\Longrightarrow\quad -3p - q + 2r = -0.35, \quad(2)\\ -27p - q + 0.15 + 8r = -3.2 &\quad\Longrightarrow\quad -27p - q + 8r = -3.35. \quad(3) \end{aligned}\]
Solving (1), (2), (3) simultaneously (using GC): \[ p = 0.2, \qquad q = 0.35, \qquad r = 0.3. \]