Another UMPT question.

**Volga** · March 13th 2011, 11:17 PM

I have this sample exam question and no model answer.

Question.

Let $Y_1,...Y_n$ be a random sample from a distribution from a density function

$f_Y(y;\theta)=\frac{2y}{\theta}e^{\frac{-y^2}{\theta}}, y>0$
$f_Y(y;\theta)=0, otherwise$

where $\theta>0$ is an unknown parameter.

i. Show that the uniformly most powerful test (UMPT) for testing hypotheses

$H_0:\theta=1, H_1:\theta>1$

will reject Ho if and only if T>K for some constant K>0, where $T=\Sigma_{i=1}^nY_i^2$ .

ii. Find the distribution of the test statistic T under Ho. [You may use the fact that the $\chi^2$ distribution with k degrees of freedom has moment generating function $(1-2t)^{-k/2}$ .

Answer.

i. The likelihood function is

$L_Y=(\frac{2}{\theta})^n\prod_{i=1}^n*y_i*e^{-\frac{1}{\theta}\Sigma_{i=1}^ny_i^2}$

The likelihood ratio from Neyman-Person lemma is

$\frac{L_Y(theta_1;y)}{L_Y(\theta_0;y)}=\frac{(\fra c{2}{\theta})^n\prod_{i=1}^ny_ie^{-\frac{1}{\theta}\Sigma_{i=1}^ny_i^2}}{2^n\prod_{i= 1}^ny_ie^{\Sigma_{i=1}^ny_i^2}}=\theta^{-n}e^{{\Sigma_{i=1}^ny_i^2(-\frac{1}{\theta}-1)}}=\theta^{-n}e^{T(-\frac{1}{\theta}-1)}$

If this ratio is higher than some number k (determined by required confidence level), then we will reject Ho:

$-\theta^{-n}e^{T(-\frac{1}{\theta}-1)}>k$

take logs

$ln(-\theta^{-n})+T(-\frac{1}{\theta}-1)>ln(k)$

$T\frac{-1-\theta}{\theta}>ln(k)-ln(-\theta^{-n})$

$T>\frac{\theta}{1+\theta}(ln(-\theta^{-n}-ln(k))=K$

and the required T>K is the above inequality.

does it make sense?

ii. Distribution of T - let me think it over a bit longer. I am thinking to find MGF of y^2 (with theta=1 the density is simplifed) and to show that it is a form of MGF of the chi squared.

**Volga** · March 14th 2011, 12:02 AM

ii. Distribution function of T

Let $Z=Y^2, T=\Sigma_{i=1}^nZ_i$

then $M_Z(t)=E(e^{Zt})=E(e^{Y^2t})=\int_0^{\infty}e^{y^2 t}2ye^{-y^2}dy=\int_0^{\infty}e^{y^2(t-1)}2ydy=\int_0^{\infty}e^{z(t-1)}dz=\frac{1}{1-t}$

(I used the fact that $\int_0^{\infty}e^{-u}du=1$ )

Then $Z_1,...Z_n$ are independent, identically distributed rvs, and therefore I can multiply their MGFs to get

$M_T(t)=(M_Z(t))^n=(\frac{1}{1-t})^n$

If I let k=2n, then this can be represented as

$(\frac{1}{1-t})^{k/2}=(1-t)^{-k/2}$

Now i need to find an error in above because this should be the 2xparameter in the expression... OR, could this still be a version of a chi-square distribution with some adjustment to [what? random variable? parameter?]

**Sambit** · March 14th 2011, 04:37 AM

Originally Posted by Volga

$-\theta^{-n}e^{T(-\frac{1}{\theta}-1)}>k$

take logs

$ln(-\theta^{-n})+T(-\frac{1}{\theta}-1)>ln(k)$

This is not allowed since log of a negative number is not defined.

Rather, write:

$-\theta^{-n}e^{T(-\frac{1}{\theta}-1)}>k$

$e^{T(-\frac{1}{\theta}-1)} < k'$ (inequality changes since multiplying by a negative number)

$T(-\frac{1}{\theta}-1) < k''$

$-T(\frac{1}{\theta}+1) < k'''$

$T > k''''$ (inequality changes since multiplying by a negative number)

Calculate k'''' from here.

NOTE: you need not write the simplified form of the constant in the RHS.

**Volga** · March 14th 2011, 04:44 AM

Thanks! Good point!

Originally Posted by Sambit

NOTE: you need not write the simplified form of the constant in the RHS.

Do you mean I don't need to work out the expression for K (as I did above), and it is sufficient to show that K is some constant?

**Sambit** · March 14th 2011, 04:47 AM

Originally Posted by Volga

Do you mean I don't need to work out the expression for K (as I did above), and it is sufficient to show that K is some constant?

Yes that is sufficient. Finally you will determine k'''' from the required confidence level; the value of k is irrelevant there; only k'''' will suffice.

**Volga** · March 15th 2011, 01:55 AM

Any help on part ii? I am still having t instead of 2t...

**Sambit** · March 15th 2011, 03:45 AM

I can't find any mistake in your solution. Your answer may not be wrong since the question does not say that the answer will involve 2t. They have only given a hint which you may or MAY NOT use. We did not use it.

**Volga** · March 15th 2011, 03:57 AM

Oh cheers! I was wondering if I completely misunderstand the MGF concept )))
then I would just say the distribution is defined by this MGF function (since it uniquely defines a distribution)

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