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: Identify the distribution of each X_k.
The density \(f_k(x)=k e^{-kx}\) for \(x>0\) is exactly the exponential density with rate parameter k. So
\[ X_1\sim \text{Exp}(1), \qquad X_2\sim \text{Exp}(2), \qquad X_3\sim \text{Exp}(3), \]
and they are given to be independent.

Step 2: Recall the minimum-of-exponentials fact.
If \(X_1,\ldots,X_m\) are independent exponential random variables with rates \(\lambda_1,\ldots,\lambda_m\), then \(Y=\min\{X_1,\ldots,X_m\}\) is again exponential, with rate equal to the sum of the individual rates. This holds because
\[ P(Y>y)=P(X_1>y)P(X_2>y)P(X_3>y)=e^{-\lambda_1 y}e^{-\lambda_2 y}e^{-\lambda_3 y}=e^{-(\lambda_1+\lambda_2+\lambda_3)y}. \]

Step 3: Apply this to our rates.
Here \(\lambda_1=1,\lambda_2=2,\lambda_3=3\), so
\[ Y\sim \text{Exp}(\lambda), \qquad \lambda=1+2+3=6. \]

Step 4: Write the moments of an exponential(6) variable.
For \(Y\sim\text{Exp}(\lambda)\),
\[ E(Y)=\frac{1}{\lambda}, \qquad \mathrm{Var}(Y)=\frac{1}{\lambda^2}, \qquad E(Y^2)=\mathrm{Var}(Y)+[E(Y)]^2=\frac{2}{\lambda^2}. \]
With \(\lambda=6\),
\[ E(Y)=\frac16, \qquad E(Y^2)=\frac{2}{36}=\frac{1}{18}. \]

Step 5: Compute E(3Y^2-Y).
By linearity of expectation,
\[ E(3Y^2-Y)=3E(Y^2)-E(Y)=3\cdot\frac{1}{18}-\frac16=\frac16-\frac16=0. \]

Final Answer:
The two terms cancel exactly. \[ \boxed{0} \]