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).
Show 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.
Step 1: Reduce the minimum to a simple exponential variable.
Let $Y_i=X_i-\mu$, which has density $g(y)=2ye^{-y^2}$ for $y>0$, free of $\mu$. Let $T=\min\{X_i\}-\mu=\min\{Y_i\}$ for $10$ such independent variables. Its survival function works out to $P(T>t)=e^{-10t^2}$.
Step 2: Square and rescale T.
Define $S=10T^2$. Then for $s>0$,
\[ P(S>s)=P\left(T>\sqrt{s/10}\right)=e^{-10(s/10)}=e^{-s} \]
So $S=10T^2$ is a standard Exponential random variable with mean $1$, a much simpler pivot to work with.
Step 3: Find the 95th percentile of this exponential pivot.
We want $s_0$ with $P(S\le s_0)=0.95$, that is $1-e^{-s_0}=0.95$, so
\[ e^{-s_0}=0.05 \quad\Rightarrow\quad s_0=\ln(20)\approx2.9957 \]
Step 4: Undo the transformation to get back to T.
Since $S=10T^2\le s_0$ with probability $0.95$, and $T>0$,
\[ T\le\sqrt{\frac{s_0}{10}}=\sqrt{0.29957}\approx0.5473 \]
So $P(0<T\le0.5473)=0.95$.
Step 5: Turn this into a confidence interval for mu.
Since $T=\min\{X_i\}-\mu$, the event $0<T\le0.5473$ rearranges to
\[ \min\{X_i\}-0.5473\le\mu<\min\{X_i\} \]
With $\min\{X_i\}=1$, the interval is $(1-0.5473,\ 1)=(0.4527,\ 1)$.
Step 6: Read off c.
\[ c\approx0.45 \]
\[ \boxed{c\approx0.45} \]
