MP test size $\alpha$ for testing $H_0:f(x)=N(0,1) \text{ vs. } H_A:f(x)=Cauchy(0,1)$ Question: Let $X$ be a single observation from a pdf $f(x)$. Find a Most Powerful test size $\alpha$ for testing $H_0:f(x)=N(0,1) \text{ vs } H_A:f(x)=Cauchy(0,1)$
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Attempt: According to the Neyman-Pearson lemma, any MP level $\alpha$ test will reject $H_0$ for large values of the likelihood ratio, which is $$\Lambda(x) = \frac{\frac{1}{\pi}\frac{1}{1 + x^2}}{\frac{1}{\root \of{2\pi}}e^{-x^2/2}} = c \frac{e^{x^2/2}}{1 + x^2} = c \frac{e^z}{1 + 2z}$$ writing $z = \frac{x^2}{2}$. We want to know for which values of $z$, $\frac{e^z}{1 + 2z}$ is larger than a threshold $k$.
Now there is a hint that was given which states:

A plot of the function $\frac{e^z}{1 + 2z}$ shows that depending on the value of $k$, the set of values of $z$ for which $\frac{e^z}{1 + 2z} > k$ is either an one sided interval of the form $z > b$ or a union of two disjoint intervals $z < a, z > b$. It is a union of two disjoint intervals if the threshold $k <1$


I did this graph on desmos to check it myself. I would have preferred it if I could have found the subject matter of the hint mathematically, as opposed to graphically looking at it.
Nonetheless, using the hint, the rejection regions of the MP test can take one of two different forms:$|X|> C$,or $|X|< A \bigcup |X|> B$
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My question is:

*

*From here, what is the MP test size $\alpha$ that completes the solution more explicity?

*Could someone please denote the rejection region(s) graphically for better understanding?

 A: How about rejecting $H_0$ if $|X|> 1.96?$
Then $\alpha = P(\mathrm{Rej}\,H_0|H_0\,\mathrm{true})= 0.05 = 5\%.$
2*pnorm(-1.96)
[1] 0.04999579

Also, $\beta = P(\mathrm{Not\,Rej}\,H_0|H_0\,\mathrm{false})=0.70.$
diff(pt(c(-1.96,1.96),1))
[1] 0.6996571

So the power of the single-observation test is about $30\%.$
In the figure below, the rejection region is outside the
vertical black lines in both tails.
hdr = "Densities of Std Norm (solid blue) and Cauchy"
curve(dnorm(x), -6,6, col="blue", lwd=2, ylab="PDF", main=hdr)
 curve(dt(x,1), add=T, col="maroon", lwd=2, lty="dotted")
 abline(h=0, col="green2"); abline(v=c(-1.96, 1.96))


Because the mean of $n=4$ observations from a standard normal
distribution has $\bar X \sim \mathsf{Norm}(\mu=0, \sigma=1/2)$
and the mean of $n$ observations from the Cauchy distribution
is again (the original) Cauchy. One can find that the critical
values $\pm c=0.980$ give $\alpha=0.05$ and $\beta = 0.494.$
q=qnorm(.025, 0, 1/2);  q
[1] -0.979982
diff(pt(c(q,-q),1))
[1] 0.4935639

hdr = "Densities of NORM(0,.5) (solid blue) and Std. Cauchy"
curve(dnorm(x,0,.5), -6,6, col="blue", lwd=2, ylab="PDF", main=hdr)
 curve(dt(x,1), add=T, col="maroon", lwd=2, lty="dotted")
 abline(h=0, col="green2"); abline(v=c(-.98, .98))


