Central Limit Theorem (normal distributions) (1 Viewer)

leehuan · May 29, 2016

I think this question may be asking me to construct a proof so I don't want to quote the CLT or anything

$$b) Let $X_1$ have a Poisson distribution with mean $\lambda_1$ and$ X_2$ with $\lambda_2$. Using the result of (a) or otherwise, show that $\\ P(X_1+X_2=n)=p_n(\lambda_1+\lambda_2)$

Previously proven:

$$Let $p_k(\lambda)=e^{-\lambda} \frac{\lambda^k}{k!}$ for $k\ge 0 \\ $be the Poisson distribution with mean $\lambda$

$$a) $\\ \sum_{k=0}^n{p_k(\lambda_1)p_{n-k}(\lambda_2}=p_n(\lambda_1+\lambda_2)$

InteGrand · May 29, 2016

leehuan said:
I think this question may be asking me to construct a proof so I don't want to quote the CLT or anything

$$b) Let $X_1$ have a Poisson distribution with mean $\lambda_1$ and$ X_2$ with $\lambda_2$. Using the result of (a) or otherwise, show that $\\ P(X_1+X_2=n)=p_n(\lambda_1+\lambda_2)$

Previously proven:

$$Let $p_k(\lambda)=e^{-\lambda} \frac{\lambda^k}{k!}$ for $k\ge 0 \\ $be the Poisson distribution with mean $\lambda$

$$a) $\\ \sum_{k=0}^n{p_k(\lambda_1)p_{n-k}(\lambda_2}=p_n(\lambda_1+\lambda_2)$

This doesn't really have anything to do with CLT.

$\noindent This result is getting us to show that the sum of two independent Poisson r.v.'s is Poisson, with parameter equal to the sum of the two original ones' parameters (and it follows from this by induction that we can extend the result to a sum of $N$ independent Poisson r.v.'s for any positive integer $N$).$

$\noindent Let $Y:= X_{1}+X_{2}$. Note that the range of $Y$ is also $0,1,2,\ldots$, since this is the range of each $X_{i}$, so the sum has this range too clearly. Now, your result in part a) has in fact given us the probability that $Y\equiv X_{1} + X_{2}$ takes on the value $n$ for any non-negative integer $n$ !! Can you explain why? If you can, this gives us the result of part b) immediately.$

leehuan · May 29, 2016

Funnily enough part c was this (I haven't attempted it yet.)

$$Let $\{X_i, 1\le i \le m \}$ be a set of $m$ random variables such that $X_i$ has a Poisson distribution with $\lambda_i. $Use the result of (b) and a proof by induction to show that the random variable$ \\ X=\sum_{i=1}^{m}{X_i}$\\ also has a Poisson distribution and that the mean is \\ $ \lambda=\sum_{i=1}^{m}{\lambda_1}$

But I guess this is the CLT part.
___________

And edit: Yep ok!

InteGrand · May 29, 2016

leehuan said:
Funnily enough part c was this (I haven't attempted it yet.)

$$Let $\{X_i, 1\le i \le m \}$ be a set of $m$ random variables such that $X_i$ has a Poisson distribution with $\lambda_i. $Use the result of (b) and a proof by induction to show that the random variable$ \\ X=\sum_{i=1}^{m}{X_i}$\\ also has a Poisson distribution and that the mean is \\ $ \lambda=\sum_{i=1}^{m}{\lambda_1}$

But I guess this is the CLT part.

This isn't CLT either. It's simply induction using previous results (similar to the Bonferoni's Inequality induction method I mentioned in another thread if you remember it).

leehuan · May 29, 2016

This was the other one I had trouble on for the time being (please)

$$Readings are rounded off to the nearest integer, and then added. Assuming the round off 'error' is uniformly distributed on $\left( -\frac{1}{2},\frac{1}{2} \right)$, use the CLT to estimate the probability that the absolute error when $n$ such readings are added exceeds $\frac{\sqrt{n}}{2}$, assuming that $n$ is not 'small'$.$

InteGrand · May 29, 2016

leehuan said:
This was the other one I had trouble on for the time being (please)

$$Readings are rounded off to the nearest integer, and then added. Assuming the round off 'error' is uniformly distributed on $\left( -\frac{1}{2},\frac{1}{2} \right)$, use the CLT to estimate the probability that the absolute error when $n$ such readings are added exceeds $\frac{\sqrt{n}}{2}$, assuming that $n$ is not 'small'$.$

$Let the error on the $i$th reading be $\varepsilon _{i}$ (this is a random variable with a uniform distribution on the given interval). Note that the overall absolute error in the sum is $\left |\sum _{i=1} ^{n} \varepsilon _{i} \right|$. Recall from uniform distribution formulas that for each $i$, we have $\mathbb{E} \left[\varepsilon _{i} \right] = 0$ and $\mathrm{Var} \left[ \varepsilon _{i} \right] = \frac{1}{12}$ (see Wikipedia link below for the mean and variance formulas for Uniform).$

$\noindent Call $Y=\sum _{i=1} ^{n} \varepsilon _{i}$, then if $n$ is large, from the CLT we have approximately $Y\sim \mathcal{N} \left(0, \sigma^2 _Y\right)$, where $\sigma ^2 _Y = n \mathrm{Var}\left[ \varepsilon _{i} \right] = \frac{n}{12}$. We want to find $\mathbb{P}\left(\left|Y\right| >\frac{\sqrt{n}}{2}\right)$. We have using this normal approximation$

$\begin{align*}\mathbb{P}\left(\left|Y\right| >\frac{\sqrt{n}}{2}\right) &= \mathbb{P}\left(Y >\frac{\sqrt{n}}{2}\right)+\mathbb{P}\left(Y <-\frac{\sqrt{n}}{2}\right) \\ &= 2 \mathbb{P}\left(Y <-\frac{\sqrt{n}}{2}\right) \quad (\text{by symmetry of a normal distribution where mean is }0) \\ &= 2 \mathbb{P}\left(Z <-\frac{\sqrt{n}}{2\sigma _{Y}}\right), \text{ where }Z \text{ is standard normal }\quad (\text{just standardising, i.e. doing }\frac{Y-\mu _Y}{\sigma_Y}\text{ and noting }\mu _Y= 0) \\ &= 2 \mathbb{P}\left(Z <-\frac{\sqrt{n}}{2\sqrt{\frac{n}{12}}}\right) \\ &= 2 \mathbb{P}\left(Z < -\frac{\sqrt{12}}{2}\right) \\ &= 2 \mathbb{P}\left(Z < -\sqrt{3}\right)\\ &\equiv 2 \Phi \left(-\sqrt{3}\right), \text{ where }\Phi \left(\cdot\right) \text{ is the standard normal cdf}. \end{align*}$

$\noindent Using a statistical table or computer software, you should find $\Phi \left(-\sqrt{3}\right) = 1 - \Phi \left(\sqrt{3}\right)\approx 0.04163226$. Hence the desired answer is approximately $2\times 0.04163226 = 0.08326452$.$

Uniform distribution: https://en.wikipedia.org/wiki/Uniform_distribution_(continuous) .

__

$\noindent To quickly see why the error in a sum is the sum of the errors, say $S_{n} = X_{1} + \cdots + X_{n}=\sum _{i=1}^{n} X_{i}$, and consider the rounded value $\widehat{S}_{n} = \sum _{i=1}^{n} \widehat{X}_{i}$, where for each $i$, $\widehat{X}_{i}$ is the rounded value of the reading or summand $X_i$, i.e. $\widehat{X}_{i} = X_{i} + \underbrace{\varepsilon_{i}}_{\in \left(-\frac{1}{2},\frac{1}{2}\right)}$. So the error in the sum is$

$\begin{align*}\widehat{S}_{n}-S_{n} &= \sum _{i=1}^{n} \widehat{X}_{i} - \sum _{i=1}^{n} X_{i} \\ &= \sum _{i=1}^{n} \left(\widehat{X}_{i} - X_{i}\right) \\ &= \sum _{i=1}^{n}\varepsilon_{i} .\end{align*}$

$\noindent To get the absolute error, just put absolute values on both sides, i.e. $\text{absolute error }\equiv \left |\widehat{S}_{n}-S_{n}\right| = \left |\sum _{i=1}^{n}\varepsilon_{i} \right | $.$

Trebla · May 30, 2016

This question is quite advanced for first year level lol

leehuan · May 30, 2016

Trebla said:
This question is quite advanced for first year level lol

Yeah a fair few of my algebra/probability questions get marked with an [H] but I still try them anyway, or want to see how they're done

mreditor16 · May 30, 2016

leehuan said:
Yeah a fair few of my algebra/probability questions get marked with an [H] but I still try them anyway, or want to see how they're done

As you will see when you start MATH1151 past papers, there will be minimal questions of [H] difficulty level in your final exam. So don't be dishheartened if nearly all of the [H] questions are close to impossible to do when you're going through tutorial hw.

integral95 · May 30, 2016

hehe if you plan to do 2nd year stats, then some of these materials are useful, otherwise you're kind of wasting your time

seanieg89 · May 30, 2016

integral95 said:
hehe if you plan to do 2nd year stats, then some of these materials are useful, otherwise you're kind of wasting your time

No way, that's such a negative mindset. Even if you don't do a followup subject, pushing yourself beyond the minimum effort required in a maths/stats subject and doing the more difficult problems will definitely sharpen your general problem solving chops. The more difficult things you get exposed to and bang your head against, the better.

leehuan · May 31, 2016

mreditor16 said:
As you will see when you start MATH1151 past papers, there will be minimal questions of [H] difficulty level in your final exam. So don't be dishheartened if nearly all of the [H] questions are close to impossible to do when you're going through tutorial hw.

Yeah I realised. Did a past paper and it wasn't the worst (though I did have trouble with wording and it took me like the whole day to get through the paper lol)
Just gotta make sure I know the required stuff.

integral95 said:
hehe if you plan to do 2nd year stats, then some of these materials are useful, otherwise you're kind of wasting your time

Pure maths major makes me pick a second year stats course

seanieg89 said:
No way, that's such a negative mindset. Even if you don't do a followup subject, pushing yourself beyond the minimum effort required in a maths/stats subject and doing the more difficult problems will definitely sharpen your general problem solving chops. The more difficult things you get exposed to and bang your head against, the better.

Yeah. I get sad over it when I don't get it but I always do it anyway because I want to progress further in my maths life than most of my peers at least lol.

Half of them come to tutorials, I tell the tutor what qns to do, then they just copy the method down as if they never did the qn before

leehuan · May 31, 2016

Getting lost again

$A varying voltage has a standard deviation of 2 volts. Its population mean value is not known and is to be estimated from the mean $\bar{X_n}$ of $n$ readings randomly taken. Use the CLT to estimate the smallest number of readings necessary to achieve a probability of 0.95 that $\bar{X_n}$ lies within 0.5V of the population mean value.$

InteGrand · May 31, 2016

leehuan said:
Getting lost again

$A varying voltage has a standard deviation of 2 volts. Its population mean value is not known and is to be estimated from the mean $\bar{X_n}$ of $n$ readings randomly taken. Use the CLT to estimate the smallest number of readings necessary to achieve a probability of 0.95 that $\bar{X_n}$ lies within 0.5V of the population mean value.$

$\noindent Let the voltage r.v. (random variable) be called $X$. Let its mean be called $\mu$ (unknown). We are given $\sigma _{X}\equiv \sigma = 2$ (the standard deviation). Let the $n$ readings be $X_{1},X_{2},\ldots, X_{n}$, then $\overline{X}_{n} = \frac{1}{n}\sum _{i=1}^{n}X_{i}$. We want the smallest $n$ such that $\mathbb{P}\left(\left | \overline{X}_{n} -\mu \right |<0.5\right)\geq 0.95$. Note that each $X_{i}$ has mean $\mu$ and standard deviation $\sigma$, so for large $n$, we have from the CLT the approximation $\overline{X}_{n} \sim \mathcal{N}\left(\mu, \frac{\sigma^2}{n}\right)$. So approximating $\overline{X}_{n} $ as this normal r.v., the standardisation process will be $Z= \frac{\overline{X}_{n} -\mu}{\frac{\sigma}{\sqrt{n}}}$, with $Z\sim \mathcal{N}\left(0,1\right)$ a standard normal random variable. Now, we have$

$\begin{align*} \mathbb{P}\left(\left | \overline{X}_{n} -\mu \right |<0.5\right) &= \mathbb{P}\left(|Z| < \frac{0.5}{\frac{\sigma}{\sqrt{n}}}\right) \quad (\text{standardisation}) \\ &= 2 \mathbb{P}\left(Z < -\frac{0.5}{\frac{\sigma}{\sqrt{n}}}\right) \quad (\text{symmetry of a standard normal pdf}) \\ &= 2\Phi \left(-\frac{0.5}{\frac{\sigma}{\sqrt{n}}}\right). \end{align*}$

$\noindent Since we wanted $\mathbb{P}\left(\left | \overline{X}_{n} -\mu \right |<0.5\right) \geq 0.95$, we equivalently want $2\Phi \left(-\frac{0.5}{\frac{\sigma}{\sqrt{n}}}\right) \geq 0.95 \Longleftrightarrow \Phi \left(-\frac{0.5}{\frac{\sigma}{\sqrt{n}}}\right) \geq 0.475.$

$\noindent Now, $\Phi \left(-\frac{0.5}{\frac{\sigma}{\sqrt{n}}}\right) = 0.475$ when $-\frac{0.5}{\frac{\sigma}{\sqrt{n}}} = \Phi ^{-1} \left(0.475\right) \equiv z_{0.475} = -0.06270678\ldots $, where $\Phi ^{-1} \left(\cdot \right)$ is the \emph{inverse CDF} (aka quantile function) of the standard normal random variable. We calculated $\Phi ^{-1} \left(0.475\right) \equiv z_{0.475} $ (which would be called the 47.5th percentile of the standard normal distribution) using a computer (alternatively, if you have a statistical table of $\Phi \left(\cdot\right)$ values, you can find the inverse cdf approximately by seeing what two values $\Phi ^{-1} \left(0.475\right) $ must be between and then using some interpolation technique like \emph{linear interpolation}).$

$\noindent Now, $-\frac{0.5}{\frac{\sigma}{\sqrt{n}}}= z_{0.475}$ when $n=\left(\frac{0.5}{\sigma \times z_{0.475} }\right)^2 = 15.89465\ldots $ (using a calculator or computer). Taking $n$ equal to this gives equality, so we attain the $\geq$ inequality by taking $n$ larger than this, because the function $\Phi \left(-\frac{0.5}{\frac{\sigma}{\sqrt{n}}}\right) $ is increasing in $n$ (to see this, recall that CDF's are increasing functions and the thing inside the CDF here is also an increasing function of $n$). So the answer is that the minimum allowable $n$ is $\boxed{n=16}$.$

InteGrand · May 31, 2016

$\noindent Here would be a way to find $z_{0.475}$ from a standard normal table. This is slightly tricky (i.e. not as trivial as just reading off a table) because such tables normally only list $\Phi \left(z\right)$ values for $z>0$, so since clearly $z_{0.475}<0$, we'll need to use symmetry properties $\Big{(}$such properties can similarly be used to find $z_{\alpha}$ from probability tables with only positive $z$ values for \emph{any} $\alpha < \frac{1}{2}$$\Big{)}$.$

$\noindent Let $z\equiv z_{0.475}$ (remember, this means $\Phi \left(z\right) = 0.475$). Then recalling that $\Phi \left(-z\right) = 1-\Phi (z)$ (symmetry property of standard normal CDF), we can take inverse CDF's of both sides to get $-z = \Phi ^{-1} \left(1-\Phi (z)\right)= \Phi ^{-1} \left(0.525\right)$. So $z\equiv z_{0.475} = - \Phi^{-1} \left(0.525\right)=-z_{0.525}$, and $0.525>0.5$, so we can estimate this from the typical standard normal CDF table.$

$\noindent In general for the standard normal percentiles, we have $z_{p}=- z_{1-p}$ for any $p \in \left(0,1\right)$.$

A typical table for the standard normal CDF is linked here: https://en.wikipedia.org/wiki/Standard_normal_table#Cumulative .

$\noindent The way to use the table to estimate percentiles is to look for $z$ values with the probability equal or nearly equal to the percentile in question. So since we are looking for $z_{0.525}$, we are looking for $z$ values there with $\Phi \left(z\right) \approx 0.525$. We see from that that $z_{0.06}=0.52392$ and $z_{0.07}=0.52790$. So the value we are after, $z_{0.525}$, must be between $0.06$ and $0.07$, by monotonicity of the inverse standard normal CDF. Using \textit{linear interpolation}, we can approximate $z_{0.525}\approx \left(\frac{0.52790-0.525}{0.52790-0.52392}\right) 0.06 + \left(\frac{0.525-0.52392}{0.52790-0.52392}\right)0.07 =0.0627 \Rightarrow z_{0.475}\approx -0.0627$.$

InteGrand · May 31, 2016

leehuan · May 31, 2016

$$One hundred and fifty plants are sprayed with a chemical which has been found 60 percent effective in providing resistance to a virus disease. Use the normal approximation to estimate the probability that at least 100 of the plants develop resistance to the virus$.$

Is this what I'm basically supposed to do (we're approximating binomial distributions)

$$Let $X$ have value 1 if resistance is developed; 0 if resistance is not developed. \\ We want $P\left(\overline{X_{150}} > \frac{100}{150} \right) \\ $ where $\mathbb E\left[\overline{X_{150}}\right]=0.6$ and Var$\left[\overline{X_{150}}\right]=\frac{0.6\times 0.4}{150}=1.6\times 10^{-3}$

$$By the CLT:$ \\ \begin{align*} P\left(\overline{X_{150}} > \frac{100}{150} \right) & \approx P\left( Z>\frac{\frac{2}{3}-0.6}{\sqrt{1.6\times 10^{-3}}} \right) \\ & = P\left(Z>1.\dot{6}\right) \\ &= 1-\Phi(1.666\dots) \\ &= 1 - \left[\frac{2}{3}\Phi(1.67) + \frac{1}{3}\Phi(1.66)\right] \\ &= 1 - \left[ \frac{2}{3}(0.9525) + \frac{1}{3}(0.9515)\right] \end{align*}$

I used the table UNSW provided for standard normal probabilities

InteGrand · May 31, 2016

Yeah that's basically what needs to be done. Although, it might be better to find Pr((100/150) < X-bar ≤ 1) instead, since X-bar can't exceed 1, but if you use the Normal approximation, you make it as though it can exceed 1, so you need to account for that by making sure you compute Pr((100/150) < X-bar ≤ 1). It'll make completely negligible numerical difference though actually in this case, so you don't need to really worry about it.

leehuan · May 31, 2016

InteGrand said:
Yeah that's basically what needs to be done. Although, it might be better to find Pr((100/150) < X-bar ≤ 1) instead, since X-bar can't exceed 1, but if you use the Normal approximation, you make it as though it can exceed 1, so you need to account for that by making sure you compute Pr((100/150) < X-bar ≤ 1). It'll make completely negligible numerical difference though actually in this case, so you don't need to really worry about it.

When X-bar = 1 I get Z=10. Lol yeah I guess it's a minute difference

InteGrand · May 31, 2016

leehuan said:
When X-bar = 1 I get Z=10. Lol yeah I guess it's a minute difference

Yeah the chance of having a normal r.v. be 10 standard deviations away from its mean is basically 0, so it makes no real difference.

Central Limit Theorem (normal distributions) (1 Viewer)

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