Discrete random variables (1 Viewer)

leehuan · May 15, 2016

Trying to calculate the variance of the uniform distribution and I don't know what the trick is to simplify the horrible looking sum

$\sigma^2\\ = \mathbb E \left[{(X-\mu)}^2\right]\\= \sum_{k=1}^n{p_k{(k-\mu)}^2}$

$$where $p_k=\frac{1}{n}$ and $\mu=\frac{n+1}{2}$

edit: wait I just realised I forgot something elementary...

InteGrand · May 15, 2016

leehuan said:
Trying to calculate the variance of the uniform distribution and I don't know what the trick is to simplify the horrible looking sum

$\sigma^2\\ = \mathbb E \left[{(X-\mu)}^2\right]\\= \sum_{k=1}^n{p_k{(k-\mu)}^2}$

$$where $p_k=\frac{1}{n}$ and $\mu=\frac{n+1}{2}$

$\noindent It will be easier to use the formula $\mathrm{Var}\left[X\right] = \mathbb{E}\left[X^2\right] - \mu^2$ here. Note that the pmf for the discrete uniform r.v. $X$ is $p_k = \frac{1}{n}$ for $k=1,2,\ldots n$ (and 0 otherwise), as you stated. Now, recall the formula $\sum _{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$. (Other formulas and generalisations may be found here:$

https://en.wikipedia.org/wiki/Faulhaber's_formula .)

$\noindent Thus$

$\begin{align*}\mathbb{E}\left[ X^2 \right] &= \sum _{k=1}^{n} k^2 \cdot \frac{1}{n} \\ &= \frac{1}{n}\times \frac{n(n+1)(2n+1)}{6} \\ &= \frac{\left(n+1\right) \left(2n+1 \right)}{6}.\end{align*}$

$\noindent So $\mathrm{Var}\left[X\right] = \mathbb{E}\left[X^2\right] - \mu^2 = \frac{\left(n+1\right)\left(2n+1\right)}{6} - \left(\frac{1+n}{2}\right)^2$. The rest is easy (but boring/tedious!) algebra.$

leehuan · May 15, 2016

InteGrand said:
$\noindent It will be easier to use the formula $\mathrm{Var}\left[X\right] = \mathbb{E}\left[X^2\right] - \mu^2$ here. Note that the pmf for the discrete uniform r.v. $X$ is $p_k = \frac{1}{n}$ for $k=1,2,\ldots n$ (and 0 otherwise), as you stated. Now, recall the formula $\sum _{k=1}^{n} = \frac{n(n+1)(2n+1)}{6}$. (Other formulas and generalisations may be found here:$

https://en.wikipedia.org/wiki/Faulhaber's_formula ).

$\noindent Thus$

$\begin{align*}\mathbb{E}\left[ X^2 \right] &= \sum _{k=1}^{n} k^2 \cdot \frac{1}{n} \\ &= \frac{1}{n}\times \frac{n(n+1)(2n+1)}{6} \\ &= \frac{\left(n+1\right) \left(2n+1 \right)}{6}.\end{align*}$

$\noindent So $\mathrm{Var}\left[X\right] = \mathbb{E}\left[X^2\right] - \mu^2 = \frac{\left(n+1\right)\left(2n+1\right)}{6} - \left(\frac{1+n}{2}\right)^2$. The rest is easy algebra.$

Yep, I basically forgot Var[X] = E[X²]-mu². But thanks anyway

leehuan · May 15, 2016

Question out of interest

Just like how you can call the Poisson distribution a limit of the binomial distribution, would you be able to say that the geometric distribution is just the binomial distribution for n=1

InteGrand · May 15, 2016

leehuan said:
Question out of interest

Just like how you can call the Poisson distribution a limit of the binomial distribution, would you be able to say that the geometric distribution is just the binomial distribution for n=1

The Bernoulli distribution is the binomial distribution with n = 1. In fact, a Bin(n,p) random variable is just a sum of n i.i.d. (independent and identically distributed) Bern(p) (Bernoulli with parameter p) random variables.

The geometric distribution can be considered a special case of the negative binomial distribution with r = 1 (or r = something else, depending on what convention is used).

Negative binomial distribution: https://en.wikipedia.org/wiki/Negative_binomial_distribution.

leehuan · May 15, 2016

Ahh right
______________________

EDIT Wow I am an idiot.. a "proportion"... (their final answer wanted alpha*n in the place of alpha basically)
$In a biochemical experiment, $n$ organisms are placed in a nutrient medium, and the number of organisms $X$ which survive for a given period is recorded. The probability distribution of $X$ is assumed to be given by $\\ P(X=k)=\frac{2(k+1)}{(n+1)(n+2)}$ for $0\le k \le n\\ $(and 0 otherwise.)$

$$b) Calculate the probability that at most a proportion $\alpha$ of the organisms survive, and deduce that for large $n$ this is approximately $a^2$

(Ignore the alpha^2 bit for now)

$Calculating $\\P(x\le \alpha) = \sum_{k=1}^\alpha{P(X=k)} \\ =\frac{2}{(n+1)(n+2)} \sum_{k=1}^{\alpha}{(k+1)}\\ =\frac{\alpha^2+3\alpha+2}{n^2+3n+2}$ is seemingly wrong$

InteGrand · May 15, 2016

leehuan said:
Ahh right
______________________

I'm doing something wrong.
$In a biochemical experiment, $n$ organisms are placed in a nutrient medium, and the number of organisms $X$ which survive for a given period is recorded. The probability distribution of $X$ is assumed to be given by $\\ P(X=k)=\frac{2(k+1)}{(n+1)(n+2)}$ for $0\le k \le n\\ $(and 0 otherwise.)$

$$b) Calculate the probability that at most a proportion $\alpha$ of the organisms survive, and deduce that for large $n$ this is approximately $a^2$

(Ignore the alpha^2 bit for now)

$Calculating $\\P(x\le \alpha) = \sum_{k=1}^\alpha{P(X=k)} \\ =\frac{2}{(n+1)(n+2)} \sum_{k=1}^{\alpha}{(k+1)}\\ =\frac{\alpha^2+3\alpha+2}{n^2+3n+2}$ is seemingly wrong$

What you are calculating is the probability that X ≤ alpha, but we need the probability that the proportion is less than or equal to alpha (whereas X is the number).

$\noindent So you'd need to find the probability that $X \leq \alpha n$. So sum the pmf from $k = 0$ to $N\left(\alpha\right)$, where $N\left(\alpha\right) \equiv \lfloor \alpha n\rfloor$ (and remember to start at $k = 0$ for this one!).$

leehuan · May 15, 2016

Hey, whatever happened to the statistics marathon thing?

InteGrand · May 15, 2016

leehuan said:
Hey, whatever happened to the statistics marathon thing?

I think a Moderator moved it a while back. It is now here: http://community.boredofstudies.org/1003/maths/350202/university-statistics-discussion-marathon.html .

leehuan · May 15, 2016

$$-The mode of the binomial probability distribution-$\\ $For the $B(n,p)$ distribution, by considering $\frac{p_k}{p_{k-1}}$, show that $p_k$ is largest when $k=\left\lfloor (n+1)p \right\rfloor$

Basically the classic HSC greatest coefficient but I've forgotten how to do simple arithmetic tonight.

$Solving $\frac{p_k}{p_{k-1}}<1$ gives $k<p(n+1)\\ $and I forgot how to use that to my advantage$

InteGrand · May 15, 2016

leehuan said:
$$-The mode of the binomial probability distribution-$\\ $For the $B(n,p)$ distribution, by considering $\frac{p_k}{p_{k-1}}$, show that $p_k$ is largest when $k=\left\lfloor (n+1)p \right\rfloor$

Basically the classic HSC greatest coefficient but I've forgotten how to do simple arithmetic tonight.

$Solving $\frac{p_k}{p_{k-1}}<1$ gives $k<p(n+1)\\ $and I forgot how to use that to my advantage$

$\noindent For the mode, $\frac{p_k}{p_{k-1}}\leq 1\Longleftrightarrow k\leq (n+1)p$ as you showed. So the $k$ with highest $p_k$ is the largest integer $k$ that is no larger than $(n+1)p$, which is precisely $\lfloor \left(n+1\right)p\rfloor$ (if $(n+1)p$ is non-integral, which is usually the case).$

$\noindent If it happens to be the case that $\left(n+1\right)p$ \emph{is} an integer, then the inequality holds iff $k \leq \left(n+1\right)p$, with \emph{attainable equality} precisely when $k=k^{\star}:=\left(n+1\right)p$ (the reason it is attainable is that $\left(n+1\right)p$ is an integer). In this case, we'll have two modes, since $p_{k^{\star}}=p_{k^{\star}-1}$ (equality case), and so $k^{\star}=\left(n+1\right)p$ and $k^{\star}-1=\left(n+1\right)p-1$ are both modes (assuming of course these are values $k$ can take on, which is an assumption we had to make for previous paragraph too in fact. So to be more exact, we could say that the mode has to be $n$ if the given modes above exceed $n$.).$

$\noindent In this analysis, we ignored the cases of $p=0$ or $q=0$ (because these should have appeared in some denominators in the analysis), where $q\equiv 1-p$ (relatively trivial cases, which would not appear much in practical settings of course). But in these cases, the mode is $0$ and $n$ respectively. This is because if $p=0$, then $p_{k} = \binom{n}{k}p^{k}q^{n-k}$ is 0 whenever $k>0$, but equals 1 when $k=0$ (using a convention of $0^0=1$, which is sometimes used in these types of settings. Also can justify this by thinking about it in terms of probability. If our probability of success on a trial is $p=0$, then we're guaranteed to get $0$ total successes, so our Binomial random variable takes value $0$ with probability 1, making $0$ the mode). Similarly we can show that the mode is $n$ if $p=1$ (i.e. $q=0$). In this case, we are guaranteed a success in each Bernoulli trial, so we get $n$ successes with probability 1, which clearly makes $n$ the mode.$

leehuan · May 16, 2016

It's always the mode.

$$Given $p_k=\frac{\lambda}{k}p_{k-1}$\\ show that $p_k$ is a maximum for $k=\alpha$ where $\lambda - 1 \le \alpha \le \lambda$

InteGrand · May 16, 2016

leehuan said:
It's always the mode.

$$Given $p_k=\frac{\lambda}{k}p_{k-1}$\\ show that $p_k$ is a maximum for $k=\alpha$ where $\lambda - 1 \le \alpha \le \lambda$

$\noindent We can't have $k=0$. Note $\frac{p_{k}}{p_{k-1}}\geq 1 \Longleftrightarrow \frac{\lambda}{k}\geq 1 \Longleftrightarrow k \leq \lambda$. So the max. $p_k$ occurs at that $k$ for which $k$ is the largest integer that is less than or equal to $\lambda$; in other words, at $k_{\text{mode}} = \lfloor \lambda \rfloor$. Equivalently, $k_{\text{mode}}$ is an integer $\alpha$ that satisfies $\lambda -1 \leq \alpha \leq \lambda$. In the case that $\lambda$ is an integer, we actually have two modes: $\lambda$ and $\lambda -1$ (provided both of these are in the range of our random variable).$

leehuan · May 16, 2016

InteGrand said:
$\noindent We can't have $k=0$. Note $\frac{p_{k}}{p_{k-1}}\geq 1 \Longleftrightarrow \frac{\lambda}{k}\geq 1 \Longleftrightarrow k \leq \lambda$. So the max. $p_k$ occurs at that $k$ for which $k$ is the largest integer satisfying that is less than or equal to $\lambda$. In other words, at $k_{\text{mode}} = \lfloor \lambda \rfloor$, or equivalently, $\lambda -1 \leq k_{\text{mode}} \leq \lambda$.$

Oops. Lol I feel dumb I was even thinking about how to manipulate p_k/p_k-1 Ok thanks again

leehuan · May 31, 2016

Never made a thread for approximations so I'll throw this question in this thread...

$Colour-blindness appears in 1 percent of the people in a certain large population. How large must a random sample be if the probability of its containing at least one colour-blind person is to be 0.95 or more? Compare exact binomial and Poisson approaches.$

InteGrand · May 31, 2016

leehuan said:
Never made a thread for approximations so I'll throw this question in this thread...

$Colour-blindness appears in 1 percent of the people in a certain large population. How large must a random sample be if the probability of its containing at least one colour-blind person is to be 0.95 or more? Compare exact binomial and Poisson approaches.$

Here's some hints

$\noindent In the Binomial model, the probability of an arbitrarily chosen person from this large population being colour-blind is $p=0.01$. If we choose a sample (independently and so on) of size $n$ which is much smaller than the total population size, this is to a good approximation `sampling with replacement', and we can model the no. of chosen people who are colour-blind $X$ as being $\text{Binomial}\left(n,p\right)$, in which case $\mathbb{P}\left(X \geq 1\right) = 1 - \mathbb{P}\left(X=0\right) = 1 - \left(0.99\right)^{n}$. From this, we can find the $n$ needed to make this probability at least $0.95$.$

$\noindent If we use a \emph{Poisson approximation}, instead we would say that if $X$ is the no. of people a sample of $n$ that have the colour-blindness, then $X \sim \text{Poisson}\left(np\right)$, so $\mathbb{P}\left(X=k\right) = e^{-{np}}\frac{\left(np\right)^{k}}{k!}$ for non-negative integers $k$. Then we would obtain $\mathbb{P}\left(X \geq 1\right) = 1 - \mathbb{P}\left(X=0\right) = 1-e^{-{n\times 0.01}}$ (as $p=0.01$). We would then use this to find the $n$ required so that this is at least $0.95$. We would then compare it to the $n$ found in the above method.$

leehuan · Jun 6, 2016

I don't believe it matters whether you have a discrete or continuous random variable so I'm randomly chucking it in this thread, but can I please have a proper example on what statistical independence is?

Like I get that Pr(A∩B) = Pr(A)Pr(B) for statistical independence but I kinda roted it, and don't fully understand it.

Which is why whilst I somewhat get this, I don't fully get it:

$If $X$ and $Y$ are statistically independent random variables, then:$
$\begin{align*} \mathbb E \left[ (X+Y)^2 \right] &=\mathbb E \left[ X^2 + 2XY + Y^2 \right] \\ &= \mathbb E \left[X^2 \right] + \mathbb E \left[2XY \right] + \mathbb E \left[Y^2 \right] \\ &= \mathbb E \left[X^2 \right] + \underbrace{2\mathbb E \left[X \right]\mathbb E \left[Y\right]}_{\text{given statistical independence}} + \mathbb E \left[Y^2 \right] \end{align*}$

Also can I have a brief refresher on why expectation is linear again? Can I just argue that because summation/integration is linear?

InteGrand · Jun 6, 2016

leehuan said:
I don't believe it matters whether you have a discrete or continuous random variable so I'm randomly chucking it in this thread, but can I please have a proper example on what statistical independence is?

Like I get that Pr(A∩B) = Pr(A)Pr(B) for statistical independence but I kinda roted it, and don't fully understand it.

Which is why whilst I somewhat get this, I don't fully get it:

$If $X$ and $Y$ are statistically independent random variables, then:$
$\begin{align*} \mathbb E \left[ (X+Y)^2 \right] &=\mathbb E \left[ X^2 + 2XY + Y^2 \right] \\ &= \mathbb E \left[X^2 \right] + \mathbb E \left[2XY \right] + \mathbb E \left[Y^2 \right] \\ &= \mathbb E \left[X^2 \right] + \underbrace{2\mathbb E \left[X \right]\mathbb E \left[Y\right]}_{\text{given statistical independence}} + \mathbb E \left[Y^2 \right] \end{align*}$

Also can I have a brief refresher on why expectation is linear again? Can I just argue that because summation/integration is linear?

$\noindent The \textit{definition} of events $A,B$ being independent is that $\mathbb{P}\left(A\cap B\right)=\mathbb{P}\left(A\right)\mathbb{P}\left(B\right)$. An intuitive `visualisation' for this is like if two coin tosses are independent, then indeed chance of two heads in a row is $\frac{1}{2}\times \frac{1}{2}$ (i.e. we can just multiply the probabilities of the individual coin tosses like done in HSC, when independence was implicitly assumed).$

$\noindent The thing done in the expectation manipulation is that if $X$ and $Y$ are two independent random variables, then the expectation of a their product is the product of their expectations. (Note this will \emph{not} generally hold if they are \emph{not} independent.) In fact, if they are independent, then $\mathbb{E}\left[f\left(X\right)g\left(Y\right)\right]=\mathbb{E}\left[f\left(X\right)\right]\mathbb{E}\left[g\left(Y\right)\right]$ for any functions $f,g$ (provided all these expectations exist). E.g. we'd also be able to say $\mathbb{E}\left[XY^3\right]=\mathbb{E}\left[X\right]\mathbb{E}\left[Y^3\right]$ if we are told $X$ and $Y$ are independent random variables. A proof of this general result (in the discrete case) may be found here:$

https://proofwiki.org/wiki/Condition_for_Independence_from_Product_of_Expectations
.

$\noindent And yes, linearity of expectation essentially comes down to linearity of summation/integration. Here's a proof in the discrete case:$

https://proofwiki.org/wiki/Linearity_of_Expectation_Function
.

$\noindent Here's a more general proof (which may not make complete sense unless you've learnt some measure theory):$

http://math.stackexchange.com/quest...r-under-what-conditions-is-median-also-a-line .

leehuan · Jul 30, 2016

Is there some way that conditional probability is supposed to work with random variables?

$\text{Assuming that the number of people alive at each age decreases in arithmetic progression, determine:}\\ \text{1. The probability a life aged 30 will survive to 50}\\ \text{2. The average future time of a 10 year old.}$

$\text{For 1. I just let }a_{30}\text{ be the amount alive at age 30 to get a final answer }\frac{a_{30}-20d}{a_{30}}\\ \text{where }d\text{ is the amount of deaths per year.}$

$\text{For 2. I got hinted to use conditional probability (which the hint may be wrong), so I started doing something like this:}\\ \text{Let }d=\text{deaths and }a=\text{initial.} \\ \begin{align*} Pr(X >10 | X=10) &= \frac{ P r( (X > 10) \cap (X=10) )}{ P r (X=10)}\\ &= \frac{ P r(?) }{\frac{a-10d}{a}} \end{align*}$

And then I got lost.

$\text{Oh, and where }X=age$

Possibly useful: My textbook (Pg 3 & Pg 14) had an example where instead they used expectation. But they actually started counting at age 0, and I have to start counting at age 10 so I wasn't sure how to use expectation here.

InteGrand · Jul 30, 2016

leehuan said:
Is there some way that conditional probability is supposed to work with random variables?

$\text{Assuming that the number of people alive at each age decreases in arithmetic progression, determine:}\\ \text{1. The probability a life aged 30 will survive to 50}\\ \text{2. The average future time of a 10 year old.}$

$\text{For 1. I just let }a_{30}\text{ be the amount alive at age 30 to get a final answer }\frac{a_{30}-20d}{a_{30}}\\ \text{where }d\text{ is the amount of deaths per year.}$

$\text{For 2. I got hinted to use conditional probability (which the hint may be wrong), so I started doing something like this:}\\ \text{Let }d=\text{deaths and }a=\text{initial.} \\ \begin{align*} Pr(X >10 | X=10) &= \frac{ P r( (X > 10) \cap (X=10) )}{ P r (X=10)}\\ &= \frac{ P r(?) }{\frac{a-10d}{a}} \end{align*}$

And then I got lost.

$\text{Oh, and where }X=age$

Possibly useful: My textbook (Pg 3 & Pg 14) had an example where instead they used expectation. But they actually started counting at age 0, and I have to start counting at age 10 so I wasn't sure how to use expectation here.

I think you would need to make an assumption about what the oldest possible age is, e.g. 100 years (or call it N years or something), because that'd clearly affect the answer.

Also, Pr(X > 10 | X = 10) is 0, so I think you meant something else.

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