Question

Let \(X_1, X_2\) and \(X_3\) be three independent random variables such that \(X_k\) \((k=1,2,3)\) has the following probability density function
\[ f_k(x)=\begin{cases} k\,e^{-kx} & \text{if } x>0 \\ 0 & \text{otherwise}. \end{cases} \]
Let \(Y=\min\{X_1,X_2,X_3\}\). Then the value of \(E(3Y^2-Y)\) equals ________ (answer in integer).

Hint:

Hint:
The minimum of independent exponential random variables with rates \(\lambda_1,\lambda_2,\lambda_3\) is exponential with rate \(\lambda_1+\lambda_2+\lambda_3\). Use this to find the moments of \(Y\).

Answer 1

Correct Answer: 0

Step 1: Write the survival function of Y from scratch.
$Y=\min\{X_1,X_2,X_3\}$ exceeds $y$ only if all three variables exceed $y$. Since $X_k$ has rate $k$, $P(X_k>y)=e^{-ky}$ for $y>0$. By independence,
\[ P(Y>y)=P(X_1>y)\,P(X_2>y)\,P(X_3>y)=e^{-y}\cdot e^{-2y}\cdot e^{-3y}=e^{-6y}. \]

Step 2: Get the density of Y.
The CDF is $F_Y(y)=1-e^{-6y}$, so differentiating gives the density
\[ g(y)=\frac{d}{dy}F_Y(y)=6e^{-6y}, \qquad y>0, \]
which is an exponential density with rate 6.

Step 3: Compute E(Y) by direct integration.
\[ E(Y)=\int_0^{\infty} y\cdot 6e^{-6y}\,dy. \]
Using the standard gamma-integral result $\int_0^{\infty} y\,e^{-\lambda y}\,dy=1/\lambda^2$ with $\lambda=6$, this becomes
\[ E(Y)=6\cdot\frac{1}{6^2}=\frac{1}{6}. \]

Step 4: Compute E(Y^2) by direct integration.
Using $\int_0^{\infty} y^2 e^{-\lambda y}\,dy=2/\lambda^3$ with $\lambda=6$,
\[ E(Y^2)=6\cdot\frac{2}{6^3}=\frac{12}{216}=\frac{1}{18}. \]

Step 5: Put the pieces together.
\[ E(3Y^2-Y)=3\Big(\frac{1}{18}\Big)-\frac{1}{6}=\frac{1}{6}-\frac{1}{6}=0. \]

Final Answer:
Building the density from the survival function and integrating directly gives the same cancellation. \[ \boxed{0} \]