Question: User: 4.2 Uniform(0,\theta)\n\nPivot T = X_{(n)}/\theta with pdf nt^{nā1} on (0,1), cdf t^n.\n\nPr(c \leq T \leq 1) = 1 ā c^n = 1 ā \alpha \to c = \alpha^{1/n}, CI: (X_{(n)}, X_{(n)} \alpha^{-1/n}).\n\nYeh shortest-length CI hai kyunki pdf increasing hai, isliye area right end (c,1) par lena best hai.\nEqual-tailed interval isse lamba hota hai.\n\nGraph 6: Uniform pivot (n = 5, \alpha = 0.05, c = 0.549)\n\nGraph shape: A smooth increasing density curve from t = 0 to t = 1, with a small left red shaded area labeled \alpha from 0 to c and a larger right blue shaded area labeled 1ā\alpha from c to 1; position_hint = center.mereko uniform kaa ci bilkul samajh nhi aa rhaa step by step mereko proper calculation do
1. **Problem Statement:**\nWe have a sample from a Uniform(0,\theta) distribution and want to find a confidence interval (CI) for \theta using the pivot \(T = \frac{X_{(n)}}{\theta}\), where \(X_{(n)}\) is the maximum order statistic.\n\n2. **Pivot Distribution:**\nThe pivot \(T\) has pdf \(f_T(t) = n t^{n-1}\) for \(t \in (0,1)\) and cdf \(F_T(t) = t^n\). This is because \(X_{(n)}\) is the maximum of \(n\) i.i.d. Uniform(0,\theta) variables.\n\n3. **Confidence Interval Construction:**\nWe want a confidence interval for \(\theta\) such that \(P(c \leq T \leq 1) = 1 - \alpha\). Using the cdf,\n$$\nP(c \leq T \leq 1) = F_T(1) - F_T(c) = 1 - c^n = 1 - \alpha.\n$$\nSolving for \(c\),\n$$\nc^n = \alpha \implies c = \alpha^{1/n}.\n$$\n\n4. **Interval in terms of \(\theta\):**\nSince \(T = \frac{X_{(n)}}{\theta}\), rearranging gives\n$$\n\theta = \frac{X_{(n)}}{T}.\n$$\nThe event \(c \leq T \leq 1\) corresponds to\n$$\n\frac{X_{(n)}}{1} \leq \theta \leq \frac{X_{(n)}}{c} = X_{(n)} \alpha^{-1/n}.\n$$\nSo the \(1-\alpha\) confidence interval for \(\theta\) is\n$$\n\left(X_{(n)}, X_{(n)} \alpha^{-1/n}\right).\n$$\n\n5. **Why shortest-length CI?**\nBecause the pdf \(f_T(t) = n t^{n-1}\) is increasing in \(t\), the highest density region is at the right end near 1. Hence, taking the interval \((c,1)\) for \(T\) gives the shortest-length confidence interval for \(\theta\).\n\n6. **Example Calculation:**\nGiven \(n=5\) and \(\alpha=0.05\), calculate \(c\):\n$$\nc = \alpha^{1/n} = 0.05^{1/5}.\n$$\nCalculate \(0.05^{1/5}\):\n$$\n\ln(c) = \frac{1}{5} \ln(0.05) = \frac{1}{5} \times (-2.9957) = -0.5991,\quad c = e^{-0.5991} \approx 0.549.\n$$\nSo the CI is\n$$\n\left(X_{(n)}, \frac{X_{(n)}}{0.549}\right).\n$$\n\n7. **Summary:**\n- Use the pivot \(T = \frac{X_{(n)}}{\theta}\) with pdf \(n t^{n-1}\).\n- Find \(c = \alpha^{1/n}\) from \(P(c \leq T \leq 1) = 1 - \alpha\).\n- The confidence interval for \(\theta\) is \(\left(X_{(n)}, X_{(n)} \alpha^{-1/n}\right)\).\n- This interval is shortest-length because the pdf is increasing, so the right tail interval is optimal.\n\nThis completes the step-by-step calculation and explanation of the uniform distribution confidence interval using the pivot method.