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TrueTears · « **Reply #1290 on:** June 28, 2012, 08:59:42 pm »

I was wondering in order to compute E(S) using the method they described, is it really as tedious as the following? (Obviously I could just apply the definition of $E(S) = \sum_{s = 2}^{12} s \times p_S(s)$ straight away, but here I wanted to try their exercise)

So in order to compute E(S) using the method they described we have to work out E(S|M=y) first which is given by $E(S|M=y) = \sum_{s = 2}^{12} s \times p_{S|M}(s|y) = \sum_{s = 2}^{12} s \left(\frac{p_{S,M}(s,y)}{p_{M}(y)}\right)$

Now we have to do that for $y = 1, \cdots, 6$ : which means we have to manually find $\sum_{s = 2}^{12} s \left(\frac{p_{S,M}(s,y)}{p_{M}(y)}\right)$ for $y = 1, \cdots, 6$

Then we have to manually find $p_M(y)$ for $y = 1, \cdots, 6$

Then after all that sub back into $E(S) = \sum_{y=1}^{6} p_M(y) E(S|M=y)$

I did all of the above and ended up with so much working, is there any shortcuts rather than applying the definitions of each expression directly?

Mao · « **Reply #1291 on:** June 30, 2012, 03:59:37 pm »

Quote from: TrueTears on June 28, 2012, 08:59:42 pm

I did all of the above and ended up with so much working, is there any shortcuts rather than applying the definitions of each expression directly?

Mathematica.

TrueTears · « **Reply #1292 on:** July 05, 2012, 12:30:51 am »

If the picture is too small, please view attached pic.

I'm just reading through the proof of the biased-ness of OLS estimates if an important independent variable is left out and I'm just stuck at the very end, basically the mathematical part that confuses me is the part circled in red.

When we take the expectation of $\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ how does it simply end as the same thing?

Ie, Why is $E\left[\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}\right] = \frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ ?

Since $\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ isn't a constant as it depends on $n$ so it's like a random variable which outputs different values as $n$ changes, so why does it follow the rule that $E(c) = c$ where $c$ is a constant?

I don't get why the part boxed in red.

Note that X, Y and U are all random variables, how is $E(\beta_1 X|X) = \beta_1 X$ ?

Shouldn't it be $E(\beta_1 X|X) = \beta_1E(X|X)$ since $\beta_1$ is just a constant, then $E(X|X) = \sum_{i=1}^n x_i P(X=x_i|X=x_i) = \sum_{i=1}^n x_i since P(X=x_i|X=x_i) = 1$

Thus $E(\beta_1 X|X) = \beta_1 \left(\sum_{i=1}^n x_i\right) \neq \beta_1 X$

Thanks

TrueTears · « **Reply #1293 on:** August 17, 2012, 03:55:54 am »

This is a pretty weird question... because:

$E(e^{tX}) = M(t) = \int_0^{\infty} e^{xt} e^{-x} dx = \int_0^{\infty} e^{-x(1-t)}dx = \lim_{k \to \infty} \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k$

But the limit: $\lim_{k \to \infty} \left[\frac{e^{k(t-1)}}{t-1}\right]$ is undefined?

How am I meant to compute the MGF then?

Thanks

Mao · « **Reply #1294 on:** August 17, 2012, 05:37:02 am »

This could be a complete misinterpretation

$M(t) = \lim_{k\to \infty} \left[ \frac{e^{k(t-1)}}{t-1} \right] - \frac{1}{t-1} = \frac{1}{1-t} \left( 1 - \lim_{k\to \infty} e^{k(t-1)}\right)$

This latter limit has two branches:

1. for $t<1$ , $e^{k(t-1)} \to e^{-\infty} = 0$ , thus $M(t) = \frac{1}{1-t}$ for $t<1$

2. for $t=1$ , $M(t=1)=0$

3. For $t>1$ , the limit diverges, thus $M(t)$ diverges.

No idea what the 'first three moments' are referring to.

TrueTears · « **Reply #1295 on:** August 17, 2012, 07:23:56 pm »

Oh right, thanks for that Mao, yup I realised that we only needed the t<1 branch, thanks again

TrueTears · « **Reply #1296 on:** September 26, 2012, 04:40:16 pm »

If we let $X_1, X_2, \cdots, X_n$ be independent and identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ then the weak law of large number states that $\bar{X_n} \rightarrow^p \mu$ and I can prove this part, however does $S^2 \rightarrow^p \sigma^2$ ? Where $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$ the sample variance? If so, how to prove it?

golden · « **Reply #1297 on:** September 26, 2012, 04:51:36 pm »

You forgot the squared for the (x-mean) part I think.

TrueTears · « **Reply #1298 on:** September 26, 2012, 05:04:30 pm »

yeah typo'ed that lol

anyways so i've tried chebyshev's inequality on:

$\lim_{n \to \infty} P(|S_n^2 - \sigma^2| \le t) \ge \lim_{n \to \infty} 1 - \frac{V(S_n^2)}{t^2}$

to somehow arrive at:

$\lim_{n \to \infty} P(|S^2 - \sigma^2| \le \epsilon) = 1$ by working backwards but can't seem to get anywhere :\

Actually I think I got it, but needed a piece of info from wiki

$V[s^2] = \sigma^4 \left( \frac{2}{n-1} + \frac{\kappa}{n} \right)$

so $\lim_{n \to \infty} P(|S_n^2 - \sigma^2| \le t) \ge \lim_{n \to \infty} 1 - \frac{V(S_n^2)}{t^2} \implies \lim_{n \to \infty} P(|S_n^2 - \sigma^2| \le t) \ge \lim_{n \to \infty} 1 - \frac{\sigma^4 \left( \frac{2}{n-1} + \frac{\kappa}{n} \right)}{t^2} \implies \lim_{n \to \infty} P(|S_n^2 - \sigma^2| \le t) =1$ and just simply let $t = \epsilon$ .

so the q is how to show $V[s^2] = \sigma^4 \left( \frac{2}{n-1} + \frac{\kappa}{n} \right)$ ? lol

golden · « **Reply #1299 on:** September 26, 2012, 05:23:45 pm »

I don't know what's going on but here's a link:
http://www.ma.utexas.edu/users/mks/M358KInstr/SampleSDPf.pdf

The part with:
We will prove that the sample variance, S2 (not MOSqD) is an unbiased estimator of the
population variance 𝜎...

Whether it is related I wouldn't know lol.

Edit: oh you seem to have it and it doesn't seem related to this haha.

TrueTears · « **Reply #1300 on:** September 26, 2012, 05:26:58 pm »

Quote from: golden on September 26, 2012, 05:23:45 pm

I don't know what's going on but here's a link:
http://www.ma.utexas.edu/users/mks/M358KInstr/SampleSDPf.pdf

The part with:
We will prove that the sample variance, S2 (not MOSqD) is an unbiased estimator of the
population variance 𝜎...

Whether it is related I wouldn't know lol.

Edit: oh you seem to have it and it doesn't seem related to this haha.

cheers for that

actually that's for another statement $E(S^2) = \sigma^2$ which is a nice result, but I was after consistency of S^2

Mao · « **Reply #1301 on:** September 27, 2012, 04:27:24 pm »

Quote from: TrueTears on September 26, 2012, 04:40:16 pm

If we let $X_1, X_2, \cdots, X_n$ be independent and identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ then the weak law of large number states that $\bar{X_n} \rightarrow^p \mu$ and I can prove this part, however does $S^2 \rightarrow^p \sigma^2$ ? Where $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$ the sample variance? If so, how to prove it?

Is the population a continuous distribution, or a large set of discrete points?

TrueTears · « **Reply #1302 on:** September 27, 2012, 04:31:43 pm »

Quote from: Mao on September 27, 2012, 04:27:24 pm

Quote from: TrueTears on September 26, 2012, 04:40:16 pm
If we let $X_1, X_2, \cdots, X_n$ be independent and identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ then the weak law of large number states that $\bar{X_n} \rightarrow^p \mu$ and I can prove this part, however does $S^2 \rightarrow^p \sigma^2$ ? Where $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$ the sample variance? If so, how to prove it?

Is the population a continuous distribution, or a large set of discrete points?

It can be any, discrete or continuous, however I am interested in the asymptotic properties

TrueTears · « **Reply #1303 on:** September 30, 2012, 10:46:48 pm »

Does anyone know the exact definition of conditional moment generating functions?

In other words, can anyone confirm whether this definition is correct or not?

Given the conditional mgf of a random variable X|Y is $m_{X|Y}(t) = f(Y;t)$ then the marginal mgf of X is given by $m_{X}(t) = E_{Y}(f(Y))$

Jenny_2108 · « **Reply #1304 on:** September 30, 2012, 10:59:34 pm »

Quote from: TrueTears on September 30, 2012, 10:46:48 pm

Does anyone know the exact definition of conditional moment generating functions?

In other words, can anyone confirm whether this definition is correct or not?

Given the conditional mgf of a random variable X|Y is $m_{X|Y}(t) = f(Y;t)$ then the marginal mgf of X is given by $m_{X}(t) = E_{Y}(f(Y))$

Hey TT, see part 3, page 20 of this link: conditional moment generating functions Hope it helps

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