ACTL1101 Questions Help (mostly first year uni probability) (1 Viewer)

leehuan · Sep 4, 2016

In the real world, is there any need to use any of the moments aside from the first (mean) and the second about the first (variance)?

(More of a personal interest than an ACTL1101 question per se)

InteGrand · Sep 4, 2016

leehuan said:
In the real world, is there any need to use any of the moments aside from the first (mean) and the second about the first (variance)?

(More of a personal interest than an ACTL1101 question per se)

Yes. The third moment has to do with skewness (which is more precisely the third standardised moment). The fourth moment has to do with kurtosis (which is more precisely the fourth standardised moment).

Skewness intuitively has to do with how asymmetric your distribution/data is about its mean, whilst kurtosis is a measure of how heavy the tails of the distribution are. These things can be important things to know about a distribution or dataset and are thus useful for example in the world of descriptive statistics.

Here are the Wikipedia pages for these for further reading:

• Skewness: https://en.wikipedia.org/wiki/Skewness
• Kurtosis: https://en.wikipedia.org/wiki/Kurtosis.

mreditor16 · Sep 4, 2016

leehuan said:
In the real world, is there any need to use any of the moments aside from the first (mean) and the second about the first (variance)?

(More of a personal interest than an ACTL1101 question per se)

Pepper your Angus for next semester. All your questions about this will be answered then haaha

leehuan · Sep 4, 2016

mreditor16 said:
Pepper your Angus for next semester. All your questions about this will be answered then haaha

Assuming I keep ACTL rip

I probably will. But it's sad how even in 1101 I'm dying

leehuan · Sep 17, 2016

Q: Suppose X follows a standard normal distribution. Calculate the correlation between X and X²

So I got lost in their working out when they computed the moments to find the covariance.
$\begin{align*}Cov(X,X^2)&=\mathbb{E}[X^3]-\mathbb{E}[X]\mathbb{E}[X^2]\\ &\stackrel{why?}{=}0\end{align*}$

This is the formula I'm allowed to use but I have a feeling for this question I'm not supposed to need it.

$\text{Standard normal: }\mathbb{E}[X^r]=\frac{1}{2^{\frac{r}{2}}} \frac{ \Gamma (1+r)}{ \Gamma \left( 1+ \frac{r}{2} \right)}$

Orthos · Sep 17, 2016

leehuan said:
Q: Suppose X follows a standard normal distribution. Calculate the correlation between X and X²

So I got lost in their working out when they computed the moments to find the covariance.
$\begin{align*}Cov(X,Y)&=\mathbb{E}[X^3]-\mathbb{E}[X]\mathbb{E}[X^2]\\ &\stackrel{why?}{=}0\end{align*}$

This is the formula I'm allowed to use but I have a feeling for this question I'm not supposed to need it.

$\text{Standard normal: }\mathbb{E}[X^r]=\frac{1}{2^{\frac{r}{2}}} \frac{ \Gamma (1+r)}{ \Gamma \left( 1+ \frac{r}{2} \right)}$

What if r is odd in that formula? What if r is even?

leehuan · Sep 17, 2016

Orthos said:
What if r is odd in that formula? What if r is even?

For the gamma function? That's not problematic

InteGrand · Sep 17, 2016

Note all odd moments are automatically 0 since X has a density that is symmetric about 0. In other words, the pdf is even, and x^r is odd if r is odd, so x^r * (pdf) is an odd function when r is odd, making the integral for the expected value 0.

leehuan · Sep 17, 2016

InteGrand said:
Note all odd moments are automatically 0 since X has a density that is symmetric about 0. In other words, the pdf is even, and x^r is odd if r is odd, so x^r * (pdf) is an odd function when r is odd, making the integral for the expected value 0.

Ahh right. Maybe I should've pulled out the integral to make it clearer for myself.

(By extension I'm guessing that the gamma function formula is only really necessary for even moments then)

InteGrand · Sep 17, 2016

$\noindent When $r$ is even, we have $r = 2k$ for some positive integer $k$. Then the moment is $\frac{\Gamma\left(1+2k\right)}{2^k \Gamma \left(1+k\right)} = \frac{(2k)!}{2^{k} k!}$. You can simplify $\frac{(2k)!}{k!}$ by the way. I'll leave that to you.$

(Don't need to worry about even moments at all for the given Q. though. Answer is just 0 due to symmetric density resulting in 0 for odd moments, as mentioned before.)

leehuan · Sep 17, 2016

InteGrand said:
$\noindent When $r$ is even, we have $r = 2k$ for some positive integer $k$. Then the moment is $\frac{\Gamma\left(1+2k\right)}{2^k \Gamma \left(1+k\right)} = \frac{(2k)!}{2^{k} k!}$. You can simplify $\frac{(2k)!}{k!}$ by the way. I'll leave that to you.$

(Don't need to worry about even moments at all for the given Q. though. Answer is just 0 due to symmetric density resulting in 0 for odd moments, as mentioned before.)

Hmm I see.

(Yeah I know)

InteGrand · Sep 17, 2016

Remark. This question gives an example of a pair of random variables (X and X^2) that are dependent yet uncorrelated. Thus uncorrelated does not imply independent. We know though that if two r.v.'s are independent, then they are uncorrelated. Thus independence is stronger than uncorrelatedness.

leehuan · Sep 17, 2016

InteGrand said:
Remark. This question gives an example of a pair of random variables (X and X^2) that are dependent yet uncorrelated. Thus uncorrelated does not imply independent. We know though that if two r.v.'s are independent, then they are uncorrelated. Thus independence is stronger than uncorrelatedness.

They mentioned that in the question

InteGrand · Sep 17, 2016

It's a pretty classic example of uncorrelated yet dependent random variables (I think it's the example given on a Wikipedia page too, and several other places, for illustrating this phenomenon.)

leehuan · Sep 22, 2016

In the real world, are there any kind of judgements needed to determine if the binomial distribution (for large n) is to be approximated by a Poisson or by a normal?

InteGrand · Sep 22, 2016

leehuan said:
In the real world, are there any kind of judgements needed to determine if the binomial distribution (for large n) is to be approximated by a Poisson or by a normal?

$\noindent Poisson approximation will only work well if $n$ is large and \textbf{$p$ is small}. This is because we recall that the result is that if $X_{n}$ is $\text{Bin}\left(n,\frac{\lambda}{n}\right)$, where $\lambda >0$ is a constant (in other words, $np$ is a constant $\lambda$), then $X_{n}$ converges to $\text{Poisson}\left(\lambda\right)$ as $n\to \infty$. (Actually we just need the $p$ for $X_{n}$ to be $\frac{\lambda_{n}}{n}$, where $\lambda_{n}\to \lambda$ as $n\to \infty$.) So for large $n$ (and hence also small $p$), $X_{n}$ will essentially behave roughly like Poisson.$

$\noindent There are probably some practical rules of thumb available for how large $n$ (and how small $p$) needs to be in order for the approximation to be useful. Note also that if $n$ is large but $p$ is also large (close to $1$), we can still make use of this result, because we can essentially just redefine what we mean by `success' in the Binomial distribution, so end up getting the small value as success probability.$

$\noindent For example, if $X$ was $\text{Bin}\left(100,0.99\right)$ say, and counting the no. of Heads in $100$ coin tosses where the coin has probability $0.99$ of showing heads each time, we couldn't approximate $X$ well with Poisson, but we could just consider the r.v. $\tilde{X}$ describing the no. of Tails obtained instead. This would be $\tilde{X}\sim \text{Bin}\left(100,0.01\right)$, and $\tilde{X}$ is related to $X$ by $\tilde{X} = 100- X$. So we can obtain any probability regarding $X$ by computing the relevant thing for $\tilde{X}$, which we \emph{can} use Poisson approximation for. (E.g. to approximate $\mathbb{P}\left( X = 87\right)$, we just need to calculate $\mathbb{P}\left(\tilde{X} = 13\right)$, which we can approximate as $e^{-\lambda}\frac{\lambda^{13}}{13!}$, where $\lambda = 100\times 0.01 = 1$.)$

$\noindent So in summary, Poisson approximation will be helpful so long as \textsl{$n$ is large and $p$ is either large or small}.$

(This result about convergence of Binomial to Poisson is known as the Poisson limit theorem, and you can read about it here: https://en.wikipedia.org/wiki/Poisson_limit_theorem .)

leehuan · Sep 26, 2016

Hypothetical scenario:

Suppose X ~ discrete r.v.
Y ~ continuous r.v.

How would you calculate Cov(X,Y)?

Also if Z ~ continuous r.v.
How would you calculate Cov(Y,Z)?

Feel free to use f_X(x) notation for pmf/pdf if need be

leehuan · Nov 3, 2016

Just briefly, how do you prove that the exponential distribution is memoryless?

InteGrand · Nov 3, 2016

leehuan said:
Just briefly, how do you prove that the exponential distribution is memoryless?

It's easy to prove that exponential => memoryless.

To prove that in fact (for continuous distribution) memoryless => exponential is a little harder – comes down to showing the only continuous function that solves the functional equation S(t+s) = S(t)S(s) will be exponential functions.

InteGrand · Nov 3, 2016

ACTL1101 Questions Help (mostly first year uni probability) (1 Viewer)

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