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In this example I have a sample out of a distribution with density $P_{\theta}(x) = 2x^{\theta} e^{−\theta x^2} I_{x \ge 0}$. I know that for all $i \ge 1$ we have $X_i^2 \sim Exp(\theta)$. If $\theta = 1$, then $T_1 = 2\sum_{1 \le i \le n}X_i^2 \sim X_{2n}^2$. And now I have to show that $T_{\theta} = 2\sum_{1 \le i \le n} \theta X_i^2$ is a pivot. A pivot is a function of which the probability function does not depend on $\theta$. But how do I show that $T_{\theta}$ does not depend on $\theta$?

Btw: I tried to plug it first in LaTeX, so that it's easier to read. But that doesn't work.

Botnakov N.
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Laura van Leuven
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  • Welcome to MSE! To learn about how to type equations properly here, see https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference. Basically you want to wrap your LaTeX codes in $ signs. E.g. typing $y = x^2$ gives $y = x^2$. – Minus One-Twelfth Jan 02 '21 at 14:05

1 Answers1

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Fix $x \ge 0$. As $X_i^2 \sim exp(\theta)$ it follow that $$P(\theta X_1^2 \le x) = P(X_1^2 \le \frac{x}{\theta }) = P( exp(\theta) \le \frac{x}{\theta })=1 - e^{- \frac{x}{\theta } \theta} = 1 - e^{-x} =P( exp(1) \le x).$$ So, we get that $\theta X_1^2 \sim \exp(1)$. Thus $Y_1 = \theta X_1^2$ is pivot. Hence $2 \sum_i Y_i$ is pivot as sum of independent pivot statistics .

Botnakov N.
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