How to Solve Continuous Random Variables — Univariate RV — moments from a continuous density Questions on Exam P
Sample Practice Problem
ID: #191KX
A filter lifetime \(X\) is uniformly distributed on \((0,8)\). It is replaced at failure or age 3, whichever comes first, so \(Y=\min(X,3)\). Calculate the variance of \(Y\).
📖 Worked Solution & Strategy
Lifetime \(X \sim \operatorname{Uniform}(0,8)\) has PDF \(f(x) = 1/8\). Observed lifetime \(Y = \min(X, 3)\) has continuous density \(1/8\) on \((0, 3)\) and a point mass \(\Pr(X > 3) = 1 - 3/8 = \frac{5}{8}\) at \(3\). The \(r\)-th moment is \(\operatorname{E}[Y^r] = \int_0^{3} x^r \left(\frac18\right)\,dx + 3^r \Pr(X > 3) = \frac{3^{r+1}}{8(r+1)} + 3^r\left(\frac{5}{8}\right)\). Evaluating gives mean \(\operatorname{E}[Y] = \frac{39}{16}\) and second moment \(\operatorname{E}[Y^2] = \frac{27}{4}\). The variance is \(\operatorname{Var}(Y) = \operatorname{E}[Y^2] - (\operatorname{E}[Y])^2 = \frac{27}{4} - (\frac{39}{16})^2 = \frac{207}{256}\). The evaluated result is \(\frac{207}{256}\), which matches the correct option.