# MATH2901 Higher Theory of Statistics (1 Viewer)

#### leehuan

##### Well-Known Member
$\bg_white f,g\text{ are increasing functions and }X,Y\text{ are i.i.d. r.v.s}$

$\bg_white \text{RTP: }(f(X) - f(Y))(g(X) - g(Y)) \ge 0$

#### InteGrand

##### Well-Known Member
$\bg_white f,g\text{ are increasing functions and }X,Y\text{ are i.i.d. r.v.s}$

$\bg_white \text{RTP: }(f(X) - f(Y))(g(X) - g(Y)) \ge 0$
We can just prove it by cases: either X ≥ Y occurs or Y > X occurs, and in either case the LHS is ≥ 0, since f and g are increasing functions.

Same is true (replacing X by x and Y by y) if x and y are just real numbers in the domain of f and g if f and g are functions defined on a subset of the reals.

#### leehuan

##### Well-Known Member
Whoops. That went over my head.

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#### He-Mann

##### Vexed?
I can't make sense out of the question.

Is it meant to be (RTP):

$\bg_white P( (f(X) - f(Y))(g(X) - g(Y))\geq 0) = 1$

#### leehuan

##### Well-Known Member
I can't make sense out of the question.

Is it meant to be (RTP):

$\bg_white P( (f(X) - f(Y))(g(X) - g(Y))\geq 0) = 1$
No, it was a function of a random variable. But InteGrand's solution worked.
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Whilst I get why chisq is the result of Z^2 (Z~N(0,1)), where in practice does it actually get used? What's so powerful about the square of the standard normal?

#### He-Mann

##### Vexed?
No, it was a function of a random variable. But InteGrand's solution worked.
The point I'm making is that how can a random variable exist alone without specification of it's probability?

Example, X is gamma random variable with whatever parameters. Prove that X > 0. How does this make sense? It should be prove that P(X > 0) = 1.

#### leehuan

##### Well-Known Member
The point I'm making is that how can a random variable exist alone without specification of it's probability?

Example, X is gamma random variable with whatever parameters. Prove that X > 0. How does this make sense? It should be prove that P(X > 0) = 1.
Look at the 2014 finals. I'm sure you still have it from back when you did the course.

I don't even care about the distribution of X in this question. I just care that X is a random variable. And I want to find something about f(X), which is what the function does to the random variable (given that f is monotonic increasing).

#### InteGrand

##### Well-Known Member
No, it was a function of a random variable. But InteGrand's solution worked.
_______________________________________

Whilst I get why chisq is the result of Z^2 (Z~N(0,1)), where in practice does it actually get used? What's so powerful about the square of the standard normal?
It's used in many hypothesis tests, for example.

#### He-Mann

##### Vexed?
Look at the 2014 finals. I'm sure you still have it from back when you did the course.

I don't even care about the distribution of X in this question. I just care that X is a random variable. And I want to find something about f(X), which is what the function does to the random variable (given that f is monotonic increasing).
Look at another simple example.

Define f: R -> R such that f(x) = x^2 and let X be a random variable with gamma distribution, say. Prove that f(X) = X^2 > 0.

It's obvious that this is true because you're just squaring positive random variables and they stay positive. But how does it make sense alone? It should be prove that P(X^2 > 0) = 1.

The point is, random variables cannot exist alone. It needs to be associated with probabilities.

#### Trebla

Whilst I get why chisq is the result of Z^2 (Z~N(0,1)), where in practice does it actually get used? What's so powerful about the square of the standard normal?
It gets used a lot to understand variances and forms the basis of other distributions such as the F-distribution (which has obvious applications in ANOVA, regression etc)

#### leehuan

##### Well-Known Member
Look at another simple example.

Define f: R -> R such that f(x) = x^2 and let X be a random variable with gamma distribution, say. Prove that f(X) = X^2 > 0.

It's obvious that this is true because you're just squaring positive random variables and they stay positive. But how does it make sense alone? It should be prove that P(X^2 > 0) = 1.

The point is, random variables cannot exist alone. It needs to be associated with probabilities.
It makes perfect sense to me after seeing the earlier question. And again, I couldn't care less if it's normal or gamma or exponential or a discrete r.v.

You've done the course before. If you're dissatisfied, pick up the exam paper and take it up with the lecturer.

It might be an abuse of notation. But otherwise I don't see what's wrong with it.

#### leehuan

##### Well-Known Member
MInd blanking.

$\bg_white f_X(x) = \alpha x^{-(\alpha+1)}, 11$

$\bg_white \text{The 80th percentile of }X\text{ is }x_{0.8}=5^{\frac{1}{\alpha}}$

$\bg_white \text{iii) Find an expression for }\mathbb{E}[X\mid X > x_{0.8}]$

#### InteGrand

##### Well-Known Member
MInd blanking.

$\bg_white f_X(x) = \alpha x^{-(\alpha+1)}, 11$

$\bg_white \text{The 80th percentile of }X\text{ is }x_{0.8}=5^{\frac{1}{\alpha}}$

$\bg_white \text{iii) Find an expression for }\mathbb{E}[X\mid X > x_{0.8}]$
By the way, this is a Pareto Distribution with scale parameter 1 (and shape parameter alpha). This distribution is related to the "80-20" law or "Pareto principle", which you may have heard of.

A fact about Pareto distributions is that the conditional distribution of a Pareto r.v. X given the event that X is greater than a given number b (where b is greater than or equal to the scale parameter of the distribution) is still a Pareto distribution, with the same shape parameter but with new scale parameter b. Using this fact and the formula for the mean of a Pareto distribution, the desired conditional mean can be deduced.

(These facts are facts that should be proved before being used for this Q. I suppose, and can be all proven as an exercise using standard methods for dealing with truncated distributions. For this particular Q., you wouldn't need to know these facts, you could just do the Q. normally, but they are interesting and it is what the Q. was probably getting at (particularly the 80-20 law, which is maybe why they chose the 80-th percentile), so I included them here.)

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#### leehuan

##### Well-Known Member
$\bg_white f_X(x) = \sqrt{\frac{\beta}{2\pi x^3}}\exp \left[-\frac{\beta}{2x}+\alpha - \frac{\alpha^2 x}{2\beta}\right]. x> 0\\ \text{and it is given that }\mathbb{E}[X] = \frac{\beta}{\alpha}$

$\bg_white \text{Some info: MLE of }\alpha\text{ is }\hat{\alpha} = \frac{\beta n}{\sum_{i=1}^n X_i}$

$\bg_white \text{The Fisher information matrix is }n \begin{pmatrix} \frac{1}{\alpha} & -\frac{1}{\beta} \\ -\frac{1}{\beta} & \frac{1}{2\beta^2}+\frac{\alpha}{\beta^2} \end{pmatrix}$

$\bg_white \text{iv) Hence show that the approximate variance for }\hat{\alpha}\text{ for large }n\text{ is}\\ \text{Var}(\hat{\alpha})=\frac{2\alpha^2+\alpha}{n}$

#### InteGrand

##### Well-Known Member
$\bg_white f_X(x) = \sqrt{\frac{\beta}{2\pi x^3}}\exp \left[-\frac{\beta}{2x}+\alpha - \frac{\alpha^2 x}{2\beta}\right]. x> 0\\ \text{and it is given that }\mathbb{E}[X] = \frac{\beta}{\alpha}$

$\bg_white \text{Some info: MLE of }\alpha\text{ is }\hat{\alpha} = \frac{\beta n}{\sum_{i=1}^n X_i}$

$\bg_white \text{The Fisher information matrix is }n \begin{pmatrix} \frac{1}{\alpha} & -\frac{1}{\beta} \\ -\frac{1}{\beta} & \frac{1}{2\beta^2}+\frac{\alpha}{\beta^2} \end{pmatrix}$

$\bg_white \text{iv) Hence show that the approximate variance for }\hat{\alpha}\text{ for large }n\text{ is}\\ \text{Var}(\hat{\alpha})=\frac{2\alpha^2+\alpha}{n}$
$\bg_white \noindent First step would be to find the 1,1 entry of the inverse of the Fisher information matrix (this would be the approximate variance of \widehat{\alpha} for large n). To eliminate \beta's from your answer, use the relationship between \widehat{\alpha} and \beta from the MLE formula, and make sure to replace \frac{1}{n}\sum_{i=1}^{n}X_{i}'s with \frac{\beta}{\alpha}, which is the mean.$

#### leehuan

##### Well-Known Member
$\bg_white \noindent First step would be to find the 1,1 entry of the inverse of the Fisher information matrix (this would be the approximate variance of \widehat{\alpha} for large n). To eliminate \beta's from your answer, use the relationship between \widehat{\alpha} and \beta from the MLE formula, and make sure to replace \frac{1}{n}\sum_{i=1}^{n}X_{i}'s with \frac{\beta}{\alpha}, which is the mean.$
For my reference sake, does the 2,2 entry of the inverse approximate Var(\hat{\beta})?

Edit: Oops.

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#### InteGrand

##### Well-Known Member
For my reference sake, does the 2,2 entry of the inverse approximate \hat{\beta}?
$\bg_white \noindent It is the approximate \textbf{variance} of \widehat{\beta} (for large n), yes.$

#### leehuan

##### Well-Known Member
Hang on, I'm not sure if I'm doing something wrong because I run into a circular argument.

$\bg_white \text{I get } \text{Var}( \hat{a} ) = \frac{n}{2\alpha\beta^2} \left( \frac{1}{2\beta^2} + \frac{\alpha}{\beta^2} \right)=\frac{n}{2\beta^4}\left( \frac1{2\alpha} +1 \right)\\ \text{which shouldn't be a problem}$

$\bg_white \text{I rearrange the definition of the MLE into }\beta = \frac{\hat{a}}{n}\sum_{i=1}^nX_i\\ \text{but then to get rid of }\sum_{i=1}^n X_i\text{ I have to put }\beta\text{ back in there?}$

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#### InteGrand

##### Well-Known Member
Hang on, I'm not sure if I'm doing something wrong because I run into a circular argument.

$\bg_white \text{I get } \text{Var}( \hat{a} ) = \frac{n}{2\alpha\beta^2} \left( \frac{1}{2\beta^2} + \frac{\alpha}{\beta^2} \right)=\frac{n}{2\beta^4}\left( \frac1{2\alpha} +1 \right)\\ \text{which shouldn't be a problem}$

$\bg_white \text{I rearrange the definition of the MLE into }\beta = \frac{\hat{a}}{n}\sum_{i=1}^nX_i\\ \text{but then to get rid of }\sum_{i=1}^n X_i\text{ I have to put }\beta\text{ back in there?}$
I think double check your Var, the beta terms should end up cancelling out I think.

#### leehuan

##### Well-Known Member
I think double check your Var, the beta terms should end up cancelling out I think.
Oh. My bad. I didn't do 1/det