Statistics Marathon & Questions (1 Viewer)

InteGrand · Jun 2, 2017

Re: Statistics

leehuan said:
Getting lost in what it means to converge in distribution to a constant

$X_n \stackrel{d}{\to} X\text{ is defined to mean}\\ \lim_{n\to \infty}F_{X_n}(x) = F_X(x)$

But I'm not seeing how F_X(x) is allowed to be a constant. Because doesn't F_X(x) technically have to satisfy these?

$F(-\infty) = 0, \quad F(\infty) = 1$

(ignoring the abuse of notation)

Converging in distribution to a constant means the limiting random variable X that we are converging in distribution to is a constant random variable (see https://en.wikipedia.org/wiki/Degenerate_distribution ). It does not mean the CDF of the limiting random variable X is constant (the CDF isn't constant in fact, of course).

leehuan · Jun 3, 2017

Re: Statistics

$\\X_1,\dots, X_n\text{ is a random sample from Unif}(0,\theta)\\ \text{Find the MLE }\hat{\theta}\text{ of }\theta$

$\\ \text{After I write down the likelihood function }L(\theta; x_1,\dots,x_n) = \prod_{i=1}^n \frac1\theta = \frac{1}{\theta^n}\text{ I'm stumped..}$

InteGrand · Jun 3, 2017

Re: Statistics

leehuan said:
$\\X_1,\dots, X_n\text{ is a random sample from Unif}(0,\theta)\\ \text{Find the MLE }\hat{\theta}\text{ of }\theta$

$\\ \text{After I write down the likelihood function }L(\theta; x_1,\dots,x_n) = \prod_{i=1}^n \frac1\theta = \frac{1}{\theta^n}\text{ I'm stumped..}$

$$\noindent The MLE is that $\widehat{\theta}$ (a function of $X_{1},\ldots,X_{n}$) that maximises that likelihood function (as a function of $\theta$). Due to what the $L$ is you've written down, the way to maximise this is to make $\widehat{\theta}$ as small as possible (since the $L$ is essentially $\theta \mapsto \frac{1}{\theta^{n}}$, which is a decreasing function of $\theta$).$

$\noindent But the smallest $\widehat{\theta}$ we can use given the sample is $\max \left\{X_{1},\ldots, X_{n}\right\}$ (why?). Thus this is the MLE.$

He-Mann · Jun 4, 2017

Re: Statistics

leehuan said:
$\\X_1,\dots, X_n\text{ is a random sample from Unif}(0,\theta)\\ \text{Find the MLE }\hat{\theta}\text{ of }\theta$

$\\ \text{After I write down the likelihood function }L(\theta; x_1,\dots,x_n) = \prod_{i=1}^n \frac1\theta = \frac{1}{\theta^n}\text{ I'm stumped..}$

Are you familiar with indicator functions?

leehuan · Jun 4, 2017

Re: Statistics

He-Mann said:
Are you familiar with indicator functions?

If I smack an indicator function onto the f(theta;x_1,...,x_n) it would be indicating on [0≤x≤theta] right?

InteGrand · Jun 4, 2017

Re: Statistics

leehuan said:
If I smack an indicator function onto the f(theta;x_1,...,x_n) it would be indicating on [0≤x≤theta] right?

$\noindent Basically, the pdf of $\mathrm{Unif}(0,\theta)$ ($\theta > 0$) isn't just $f(x) = \frac{1}{\theta}$, it's $f(x) = \frac{1}{\theta}I_{[0,\theta]}(x)$. Therefore, given non-negative data $x_{1},x_{2},\ldots, x_{n}$, we have $f(x_{i}; \theta) = \frac{1}{\theta}I_{[0,\theta]}(x_{i})$. So the likelihood is the product of these from $i=1$ to $n$, i.e. it is $\frac{1}{\theta^{n}}\prod_{i=1}^{n}I_{[0,\theta]}(x_{i})$. Now, since $\prod_{i=1}^{n}I_{[0,\theta]}(x_{i}) = I_{[0,\theta]}\left(x_{(n)}\right)$, where $x_{(n)} = \max \left\{x_{1},\ldots, x_{n}\right\}$ (easy exercise), the likelihood is $\frac{1}{\theta^{n}} I_{[0,\theta]}\left(x_{(n)}\right)$. This is maximised as a function of $\theta$ for fixed data at $\theta = x_{(n)}$. This explains why the MLE is the max. of the data.$

leehuan · Jun 4, 2017

Re: Statistics

$\text{I have a joint distribution}\\ f_{(X_1,X_2)}(x_1,x_2) = \frac{1}{2\sqrt{3}\pi} e^{ -\frac12 \left[ \frac{(x_1-4)^2}{3} +(x_2-2)^2 \right] } \text{ for } x\in \mathbb{R},y \in \mathbb{R}$

Can I just assume that I have a bivariate Gaussian distribution with X₁ and X₂ independent?

InteGrand · Jun 4, 2017

Re: Statistics

leehuan said:
$\text{I have a joint distribution}\\ f_{(X_1,X_2)}(x_1,x_2) = \frac{1}{2\sqrt{3}\pi} e^{ -\frac12 \left[ \frac{(x_1-4)^2}{3} +(x_2-2)^2 \right] } \text{ for } x\in \mathbb{R},y \in \mathbb{R}$

Can I just assume that I have a bivariate Gaussian distribution with X₁ and X₂ independent?

Yes, this is the case.

Flop21 · Jun 6, 2017

Re: Statistics

For finding the var of this, why in the answers do they square 1/n ? Isn't it just 1/n sum(i=0..n) Var (Xi)??

Flop21 · Jun 6, 2017

Re: Statistics

To find the variance of this.... why do the answers have 4Var(X), where does this 4 come from??

Thanks

InteGrand · Jun 6, 2017

Re: Statistics

Flop21 said:
For finding the var of this, why in the answers do they square 1/n ? Isn't it just 1/n sum(i=0..n) Var (Xi)??

Flop21 said:
To find the variance of this.... why do the answers have 4Var(X), where does this 4 come from??

Thanks

Remember, there is an identity Var(cX) = c² Var(X), if c is a constant. In other words, to get the variance of a constant times something, the constant can come outside but gets squared.

leehuan · Jun 6, 2017

Re: Statistics

Does sure convergence necessarily imply every other form of convergence?

Flop21 · Jun 7, 2017

Re: Statistics

Finding the confidence interval of something, but where do they get the variance for the se in this picture???

InteGrand · Jun 7, 2017

Re: Statistics

Flop21 said:
Finding the confidence interval of something, but where do they get the variance for the se in this picture???

It's a standard formula for the variance of a sample proportion, and basically comes from the formula for the variance of a Bernoulli(p) random variable (or variance of a Binomial(n, p) random variable).

It's worked through on this page, for example: http://amsi.org.au/ESA_Senior_Years/SeniorTopic4/4g/4g_2content_3.html (scroll down a bit to the section "Mean and variance of the sample proportion" for the derivation part).

Flop21 · Jun 7, 2017

Re: Statistics

InteGrand said:
It's a standard formula for the variance of a sample proportion, and basically comes from the formula for the variance of a Bernoulli(p) random variable (or variance of a Binomial(n, p) random variable).

It's worked through on this page, for example: http://amsi.org.au/ESA_Senior_Years/SeniorTopic4/4g/4g_2content_3.html (scroll down a bit to the section "Mean and variance of the sample proportion" for the derivation part).

Thank you this helps so much!

Flop21 · Jun 7, 2017

Re: Statistics

Why are the answers using Xbar +- 1.96 sqr(Xbar/n) for confidence intervals?? With 95% confidence.

I can only find formulas for Xbar +- Z * delta/sqr(n)

And I thought Z = 1-0.05/2 for 95% confidence...

--

Sorry if stupid questions, but I'm so lost with this course, super behind. And I can't seem to find the right formulas in the notes.

InteGrand · Jun 7, 2017

Re: Statistics

Flop21 said:
Why are the answers using Xbar +- 1.96 sqr(Xbar/n) for confidence intervals?? With 95% confidence.

I can only find formulas for Xbar +- Z * delta/sqr(n)

And I thought Z = 1-0.05/2 for 95% confidence...

--

Sorry if stupid questions, but I'm so lost with this course, super behind. And I can't seem to find the right formulas in the notes.

$\noindent It would be because the distribution of the test statistic is (or approximately is) normal, so a $95\%$ confidence interval would be the estimate plus-or-minus $1.96$ times its standard error. This comes from the fact that a standard normal distribution has 97.5-th percentile point equal to 1.96 (approximately). That is, if $X\sim \mathcal{N}(\mu,\sigma^{2})$ (i.e. $X$ is a normal random variable with mean $\mu$ and variance $\sigma^{2}$), we have that $X$ falls between $\mu -1.96 \sigma$ and $\mu + 1.96 \sigma$ with probability $95\%$ (approximately). There's a Wikipedia article on 1.96:$

https://en.wikipedia.org/wiki/1.96 .

He-Mann · Jun 7, 2017

Re: Statistics

LOL! Wikipedia article on '1.96'.

Flop21 · Jun 11, 2017

Finding the maximum likelihood estimator of theta when X1, X2...Xn is a random sample from the density:

fX(x;theta) = theta * e^(-theta*x), x > 0; theta > 0.

So to do this I tried, finding the likelihood estimator, then the log likelihood estimator getting... sum(i=0..n)[ln(theta * e^(-theta*x))].

Then simplifying to.... nln(theta) - theta * sum(x). Then differentiating i got, n/theta - sum(x)... theta then = n/sum(x).

But the answers show, 1/X(bar). How did they get this?

InteGrand · Jun 11, 2017

Flop21 said:
Finding the maximum likelihood estimator of theta when X1, X2...Xn is a random sample from the density:

fX(x;theta) = theta * e^(-theta*x), x > 0; theta > 0.

So to do this I tried, finding the likelihood estimator, then the log likelihood estimator getting... sum(i=0..n)[ln(theta * e^(-theta*x))].

Then simplifying to.... nln(theta) - theta * sum(x). Then differentiating i got, n/theta - sum(x)... theta then = n/sum(x).

But the answers show, 1/X(bar). How did they get this?

$\noindent Those are the same. Since $\overline{X} = \frac{\sum\limits_{i=1}^{n}X_{i}}{n}$ (definition of the sample mean), we have $\frac{1}{\overline{X}} = \frac{n}{\sum\limits_{i=1}^{n}X_{i}}$.$

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