Question

Let \(X_1,X_2,\ldots,X_{10}\) be a random sample from the following probability density function
\[f(x)=\begin{cases}2(x-\mu)e^{-(x-\mu)^2} & \text{if } x>\mu\\0 & \text{otherwise,}\end{cases}\]
where \(\mu\in(-\infty,\infty)\) is an unknown parameter. It is given that the observed value of \(\min\{X_1,X_2,\ldots,X_{10}\}\) is \(1\). Using the pivot \(\min\{X_1,X_2,\ldots,X_{10}\}-\mu\), suppose a 95% confidence interval of \(\mu\) is of the form \((c,1)\), then \(c\) equals ______ (rounded off to two decimal places).

Hint:

Hint:
Show that \(Y=X-\mu\) has distribution function \(1-e^{-y^2}\), find the survival function of \(\min\{X_i\}-\mu\), and use it as a pivot to build a one-sided confidence interval.

Answer 1

Correct Answer: 0.45

Step 1: Simplify the density with a shift.
Let \(Y_i=X_i-\mu\). Since \(f(x)=2(x-\mu)e^{-(x-\mu)^2}\) for \(x>\mu\), the variable \(Y_i\) has density
\[ g(y)=2y\,e^{-y^2}, \qquad y>0 \]
which does not involve \(\mu\) at all. So the shape of \(Y_i\) is fixed and known.

Step 2: Find the distribution function of Y.
Integrating the density,
\[ G(y)=\int_0^y 2t\,e^{-t^2}\,dt=\left[-e^{-t^2}\right]_0^y=1-e^{-y^2}, \qquad y>0 \]

Step 3: Find the distribution of the minimum.
Let \(T=\min\{X_1,\dots,X_{10}\}-\mu=\min\{Y_1,\dots,Y_{10}\}\). Since the \(Y_i\) are independent copies of \(Y\),
\[ P(T>t)=\prod_{i=1}^{10}P(Y_i>t)=\big(1-G(t)\big)^{10}=\big(e^{-t^2}\big)^{10}=e^{-10t^2}, \qquad t>0 \]
Because this survival function is free of \(\mu\), \(T\) is a valid pivotal quantity.

Step 4: Build a one-sided 95% interval for the pivot.
Since \(T>0\) always, we look for \(b>0\) such that
\[ P(0<T\le b)=0.95 \quad\Longleftrightarrow\quad P(T>b)=0.05 \]
Using the survival function from Step 3,
\[ e^{-10b^2}=0.05 \]

Step 5: Solve for b.
\[ -10b^2=\ln(0.05) \quad\Rightarrow\quad b^2=\frac{-\ln(0.05)}{10}=\frac{\ln(20)}{10} \]
Since \(\ln(20)\approx2.9957\),
\[ b^2\approx0.29957 \quad\Rightarrow\quad b\approx0.5473 \]

Step 6: Convert the pivot interval into an interval for mu.
The event \(0<T\le b\) is the same as
\[ 0<\min\{X_i\}-\mu\le b \quad\Longleftrightarrow\quad \min\{X_i\}-b\le\mu<\min\{X_i\} \]
So the 95% confidence interval is \(\big(\min\{X_i\}-b,\ \min\{X_i\}\big)\), matching the required form \((c,1)\) since the observed \(\min\{X_i\}=1\).

Step 7: Compute c.
\[ c=1-b\approx1-0.5473=0.4527 \]

Final Answer:
Rounded to two decimal places, \[ \boxed{c\approx0.45} \]