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S Jun 15, 2023 at 16:14 history suggested Starship CC BY-SA 4.0
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S Jun 15, 2023 at 16:14
Dec 14, 2017 at 23:04 comment added Gordon Smyth One last hint. $\sum 2\theta X^2_i$ follows a chi-square distribution, but not on $n$ df. You figure out the df.
Dec 14, 2017 at 18:58 comment added Gordon Smyth $2\theta X^2_i$ is obviously a strictly positive quantity. I wonder how you expected to show that a strictly positive random variable follows a $N(0,1)$ distribution? That should have been ringing alarm bells for you.
Dec 14, 2017 at 18:28 comment added Gordon Smyth Pivoting and normality have nothing at all to do with one another. A pivot is just a random variable whose distribution doesn't depend on $\theta$. There is no requirement for it to follow a normal, or any other, distribution. (Surely you could have found out for yourself, either from the course notes or from google en.wikipedia.org/wiki/Pivotal_quantity .)
Dec 14, 2017 at 1:24 vote accept Silvia Rossi
Dec 14, 2017 at 0:48 answer added Deep North timeline score: 1
Dec 14, 2017 at 0:40 comment added Silvia Rossi Thank you very much, I will do that. But how would I show that $\sum 2 \theta X_i^2$ is a pivot if it does not have $N(0,1)$ distribution? @GordonSmyth
Dec 13, 2017 at 23:55 comment added Gordon Smyth Just apply the transformation $y=x^2$. You will find that $y$ has a very simple well-known distribution. You will not however be able to show that $\sum 2\theta X_i^2\sim N(0,1)$ because it is simply not true.
Dec 13, 2017 at 23:16 history edited Silvia Rossi CC BY-SA 3.0
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Dec 13, 2017 at 22:50 history edited Silvia Rossi CC BY-SA 3.0
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Dec 13, 2017 at 22:49 history edited Taylor CC BY-SA 3.0
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Dec 13, 2017 at 22:37 history edited Silvia Rossi
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Dec 13, 2017 at 22:35 comment added Michael R. Chernick Add the self study tag.
Dec 13, 2017 at 22:29 history asked Silvia Rossi CC BY-SA 3.0