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The following is a proof of the validity of the (bootstrap) percentile interval taken from Larry Wasserman's "All of Statistics". $\theta^*_{\alpha/2}$ and $\theta^*_{1-\alpha/2}$ denote the respective quantiles of the bootstrap of our estimator $\hat {\theta}$.

Could you please take a look at the proof below and explain to me where the third line of the derivation comes from? How does $U$ enter the derivation? Is there perhaps a mistake there?

Thank you in advance.

enter image description here

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  • $\begingroup$ @COOLSerdash Would you mind telling me how you enlarged the picture? For future reference ;) $\endgroup$
    – JohnK
    Jun 14, 2014 at 10:25
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    $\begingroup$ Hi John. I just saved the picture and opened it in a graphics editing program (I used paint.NET but Windows Paint would also suffice). I then selected the relevant area and selected "crop to selection" so that all the white margins were cut off. Then I saved the picture again and uploaded it. $\endgroup$ Jun 14, 2014 at 10:41

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I think the source of confusion is the in the two sentences before the derivation. Before the "Hence", I would add: "Suppose also that conditional on $U$, $U^*_b \sim N(U, c^2)$." Then, the next sentences should be, "Hence, $u_{\alpha/2}^* = U - z_{\alpha/2} c$ and $u_{1-\alpha/2}^* = U + z_{\alpha/2} c$."

A transformation $m$ that satisfies both the original "Suppose" and my addition will rarely exist. However, it will usually exist asymptotically as the sample size grows. This is what Wasserman is alluding to in his last sentence. Essentially, $m$ existing means that the true distribution of your estimator, $\hat{\theta}$, and the bootstrap distribution of $\theta^*_{n,b}$ are the same.

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  • $\begingroup$ Thank you, I find his book packed with that kind of nonsense. $\endgroup$
    – JohnK
    Jun 20, 2014 at 19:10

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