Consider the most powerful test of the null hypothesis that $X$ is a standard normal random variable against the alternative that $X$ is a random variable having pdf

$$f(x) =\frac{2}{\Gamma(1/4)}\text{exp}(−x^4)\text{ }I_{(-\infty,\infty)}(x)$$

and give the p-value if the observed value of $X$ is $0.6$

My try:

I think I should be using a likelihood ratio test.

I read that the Neyman–Pearson lemma states that the likelihood ratio test is the most powerful among all level $\alpha$ tests.

We have that the likelihood ratio is


Thus we accept $H_0$ if

$$\frac{\Gamma(1/4)}{\sqrt{8\pi}}\text{exp}\left(\frac{-x^2}{2}+x^4\right)\geq c$$

or equivalently if

$$\frac{-x^2}{2}+x^4 \geq \text{log}\left(\frac{\sqrt{8\pi}\cdot c}{\Gamma(1/4)}\right)$$

or equivalently if one of the following holds:

$$x^2\geq \frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}$$

$$x^2\geq \frac{1-\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}$$

or equivalently if one of the following holds:

$$x\geq \sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}$$

$$x\leq -\sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}$$

$$x\geq \sqrt{\frac{1-\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}$$

$$x\leq -\sqrt{\frac{1-\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}$$

For a meaningful acceptance region we only consider the top two of the four constraints. Hence we reject if $$x\in\left(-\sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}},\sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}\right)$$

We wish, under the null, that the probability that $X$ assumes a value in this range to be $0.05$

For this to occur, we need

$$\sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}=0.06270678$$

But software gives that there are no solutions for $c\in\mathbb{R}$. Any suggestions or confirmation of my approach would be much appreciated.

  • $\begingroup$ One of the roots of the bi-quadratic equation is negative because $c\ge\frac{\Gamma(1/4)}{\sqrt(8 \pi)}$, therefore you only will have the condition that |x|>... $\endgroup$
    – sega_sai
    Commented Dec 2, 2018 at 1:52
  • $\begingroup$ Am I correct in saying that it must be that $0\leq 1+16\cdot\text{log}(1.382746\cdot c)\leq 1$ and so $c\in(0.679,0.723)$? $\endgroup$
    – Remy
    Commented Dec 2, 2018 at 1:58
  • $\begingroup$ No. $1\le1+16\log(\frac{\Gamma(1/4)}{\sqrt{2\pi}} c)$ for all meaningful c. $\endgroup$
    – sega_sai
    Commented Dec 2, 2018 at 2:31
  • $\begingroup$ That would make the bottom two of my four constraints undefined so should I only be concerned with the top two? $\endgroup$
    – Remy
    Commented Dec 2, 2018 at 2:34
  • 1
    $\begingroup$ You are doing something incorrect here. To find out the c for a given p-value, you need to consider the distribution of you statistic under null -- in this case your TS is $\frac{\Gamma(1/4)}{\sqrt{8 \pi}}\exp(-x^2/2+x^4)$ and find out what value corresponds to 0.05/0.95 tail probality. That's what you need to use as a $c$. But the answer below by StubbornAtom is the right one anyway. $\endgroup$
    – sega_sai
    Commented Dec 2, 2018 at 7:59

2 Answers 2


Possible way to proceed:

By N-P lemma, a most powerful test of size $\alpha$ for testing $H_0$ against $H_1$ is the indicator function


, where $$\lambda(x)=\frac{f_{H_1}(x)}{f_{H_0}(x)}$$ and $k$ is so chosen that $$E_{H_0}\varphi(X)=\alpha$$



Or, $$\ln \lambda(x)=\frac{x^2}{2}-x^4+\text{ constant }$$

Log being monotonically increasing, it suffices to study the nature of this function. You can consider this to be a function of $x^2$, plot the function, differentiate and check the sign of the derivative. We are interested to know whether $\lambda(x)$ is increasing or decreasing in $x$, or both increasing and decreasing for different values of $x$.

Ultimately you would have to reach some conclusion regarding $x$ from $\lambda(x)$ if possible, like

$$\lambda(x)>k\implies k_1<x^2<k_2\implies c_1<x<c_2$$

Apart from possible relations between $c_1,c_2$ which can be found through the analysis above, additional constraint on $c_1,c_2$ is of course the size (or level) restriction.

The critical region gets complicated if you go into details.

For $\lambda(x)>k$, we have for some $k_1$,

\begin{align} \frac{x^2}{2}-x^4&>k_1 \\\implies x^4-2x^2\cdot\frac{1}{4}+\frac{1}{16}&<\frac{1}{16}-k_1 \end{align}

Or, $$\left(x^2-\frac{1}{4}\right)^2<\frac{1}{16}-k_1$$

So for some $c$, the critical region is the set of all $x$ such that

\begin{align} \left|x^2-\frac{1}{4}\right|<c \iff -c+\frac{1}{4}<x^2<c+\frac{1}{4} \end{align}

You can stop here. If you want you can solve this further for $x$, but not necessary I think.

If you can find $c$ such that $P_{H_0}(-c+\frac{1}{4}<X^2<c+\frac{1}{4})=\alpha$, then a level $\alpha$ MP test is ready using N-P lemma. You know that under $H_0$, $X^2$ has a $\chi^2_1$ distribution.

  • 1
    $\begingroup$ Thank you, I will reattempt with this in mind and edit my post $\endgroup$
    – Remy
    Commented Dec 1, 2018 at 21:50
  • $\begingroup$ Just to make sure, we want $\mathsf P\left(\frac{\sqrt{8\pi}}{\Gamma(1/4)}\text{exp}\left(\frac{x^2}{2}-x^4\right)\gt k\right)=0.05$? $\endgroup$
    – Remy
    Commented Dec 1, 2018 at 23:00
  • $\begingroup$ So we reject the null hypothesis if our observation squared falls between $-c+\frac{1}{4}$ and $c+\frac{1}{4}$? And would the desired p-value then just be the probability of a chi-square random variable taking on a value less than $0.36$? $\endgroup$
    – Remy
    Commented Dec 2, 2018 at 8:04
  • $\begingroup$ @Remy I am afraid you need expert opinion on that. The purpose of my answer was to show how the critical region could be simplified using N-P lemma. If my test statistic is $T=X^2$, observed value of $T$ is $t=0.6^2=0.36$. I am not sure what exactly is the p-value in this setup. (It could be $P_{H_0}(T\ge t)$ or $P_{H_0}(T\le t)$ or twice the minimum of the two probabilities.) If the level is $\alpha$ (known), then you are to reject $H_0$ if the p-value is less than or equal to $\alpha$. That is, if you have to make a decision rule based on p-value. $\endgroup$ Commented Dec 2, 2018 at 11:06

The test will reject $H_0$ for sufficiently large values of the ratio

$$\begin{align*} \frac{2}{\Gamma\left(\frac{1}{4}\right)}\frac{\text{exp}\left(-x^4\right)}{\frac{1}{\sqrt{2\pi}}\text{exp}\left(-\frac{x^2}{2}\right)} &=\frac{2\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)}\text{exp}\left(-x^4+\frac{1}{2}x^2\right)\\\\ &=\frac{2\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)}e^{\frac{1}{16}}\text{exp}\left(-x^4+\frac{1}{2}x^2-\frac{1}{16}\right)\\\\ &=c\cdot\text{exp}\left(-\left[x^2-\frac{1}{4}\right]^2\right) \end{align*}$$

where $c$ is a positive constant. For the observed value of $x$, this ratio equals


The desired p-value is the probability, when $H_0$ is true, that $X$ assumes any value $x$ such that $$c\cdot\text{exp}\left(-\left[x^2-\frac{1}{4}\right]^2\right)\geq c\cdot\text{exp}(-0.0121)$$

Such values of $x$ satisfy

$$\left[x^2-\frac{1}{4}\right]^2\leq 0.0121\Rightarrow\left|x^2-\frac{1}{4}\right|\leq0.11$$

Hence $$-0.11\leq x^2-\frac{1}{4}\leq0.11\Rightarrow 0.14\leq x^2 \leq 0.36$$

Then $$\sqrt{0.14}\leq x \leq 0.6 \text{ or } -0.6\leq x \leq -\sqrt{0.14}$$

Because the standard normal pdf is symmetric about $0$, the desired p-value is



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