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Let $y_1,...,y_n$ be a random sample from a normal distribution, $N(\theta,\theta)$, where $\theta>0$.

Show that $$U=\frac{\overline{Y}_n-{\theta}}{\sqrt{\frac{\theta}{n}}}$$ is a pivotal quantity where $\overline{Y}_n$ is the sample mean.

I am stuck on property 2 of the pivotal quantity. So, how do I show that the probability distribution of $U$ does not depend on $\theta$?

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    Show $U \sim N(0,1)$ with parameters unrelated to $\theta$ – 2017-02-21

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Since $U$ is a linear combination of Gaussian variables, it is a Gaussian variable. We only need to show that the expectation of $U$ and the variance of $U$ do not depend on $\theta$.

  • The expectation of $y_i$ is $\theta$ for all $i$. Due to the linearity of the expectation, the expectation of $U$ is $E(U) = \frac{1}{\sqrt{\theta/n}} \left(\frac{1}{n}\,n\theta - \theta\right) = 0$ .
  • The variance of $y_i$ is $\theta$ for all $i$. Due to the scaling and summation properties of the variance of uncorrelated variables, the variance of $U$ is $V(U) = \frac{n}{\theta} \left(\frac{1}{n^2}\,n\theta - 0\right) = 1$ .

Thus, $U\sim\mathcal{N}(0,1)$ is a pivotal quantity.

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    I see, thank you! What if instead of $\sqrt{\theta}$ we have $S$, the sample variance? Do I still show that $U$~$N(0,1)$? – 2017-02-21
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    oops, I meant S, the sample standard deviation, where http://essedunet.nsd.uib.no/cms/topics/multilevel/img/variance.PNG – 2017-02-21
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    @sucksatmath, if you have $S$ instead of $\sqrt\theta$, you get $U\sim T(n-1)$ Student-T distibution with $n-1$ degrees of freedom. – 2017-02-22