23:I[90314,["https://schon.au/_next/static/immutable/chunks/3w1yo8ttdl7xf.js","https://schon.au/_next/static/immutable/chunks/1t10en_fq84vd.js","https://schon.au/_next/static/immutable/chunks/3cp788yseft7o.js","https://schon.au/_next/static/immutable/chunks/1em42ny1s8_11.js","https://schon.au/_next/static/immutable/chunks/44ry3tv25aq6z.js","https://schon.au/_next/static/immutable/chunks/1k90lc_78gxlx.js"],"default"] 24:I[86022,["https://schon.au/_next/static/immutable/chunks/3w1yo8ttdl7xf.js","https://schon.au/_next/static/immutable/chunks/1t10en_fq84vd.js","https://schon.au/_next/static/immutable/chunks/3cp788yseft7o.js","https://schon.au/_next/static/immutable/chunks/1em42ny1s8_11.js","https://schon.au/_next/static/immutable/chunks/44ry3tv25aq6z.js","https://schon.au/_next/static/immutable/chunks/1k90lc_78gxlx.js"],"default"] 25:T351b,MATLAB `disttool` for distributions. ### 3.1 The Binomial Distribution Use **binomial distribution** when: - The outcome of a random experiment can be classified as either a "Success" or "Failure" - We observe a "Success" with probability $\pi$ - $n$ **independent** repetitions of this experiment are performed Define $X$ to be the number of "Successes" observed over $n$ repetitions. We say that $X$ is a **binomial random variable** with parameters $n$ and $\pi$, i.e. $X\sim\text{Bin}(n,\pi)$. The binomial probability mass function is given by $$p(x)=\displaystyle\binom{n}{x}\pi^x(1-\pi)^{n-x}$$ **Binomial Distribution Properties:** - Using the binomial expansion, $$\displaystyle \sum_{x \in S_X} p(x) = \sum_{x=0}^n \binom{n}{x}\pi^x(1-\pi)^{n-x} = (\pi+(1-\pi))^n =1$$ - If $X_1 \sim \text{Bin}(n_1,\pi)$ and $X_1$ is **independent** of $X_2$, then, $$X_1+X_2\sim\text{Bin}(n_1+n_2,\pi)$$ If $X\sim\text{Bin}(n,\pi)$: - ==Mean==: $\mu = \mathbb{E} (X) = n \pi$ - ==Variance==: $\sigma^2=\mathbb{V}\operatorname{ar}(X)=n\pi(1-\pi)$ > [!EXAMPLE] Example: Binomial Distribution > It is known that disks produced by a certain company will be defective with probability 0.01 independently of each other. The company sells the disk in packages of 10 and offers a money-back guarantee if more than 1 of the disks are defective. > *** > 1. In the long-run, what proportion of package purchases can get their money back? > $$\begin{align*} > P(X\gt 1)&=1-P(X\leq 1)\\ > &=1-P(X=0)-P(X=1)\\ > &=1-\pmatrix{10\\0}0.01^0(1-0.01)^{10}-\pmatrix{10\\1}0.01^1(1-0.01)^{10-1}\\ > &=1-0.99^{10}-10\times0.01\times0.99^9\\ > &\approx0.0043 > \end{align*}$$ > 2. If someone buys three packages, what is the probability that they could get their money back on exactly one of them? > $$\begin{align*} > \text{Let }Y&\text{ be the number returned, out of }3\\ > Y\sim&\text{Bin}(3,0.0043)\\\\ > P(Y=1)&=\pmatrix{3\\1}0.0043^1(1-0.0043)^{3-1}\\ > &\approx0.013 > \end{align*}$$ > > **OR Using MATLAB:** > ```matlab > p=1-0.99^10-10*0.01*0.99^9 > > 1-binocdf(1,10,0.01) > binopdf(1,3,p) > ``` ### 3.2 The Poisson Distribution We use the Poisson distribution the **number of occurrences** of some **random phenomenon** in a **fixed period of time**. Suppose $X$ is the number of occurrences. We say that $X$ is a Poisson random variable with parameter $\lambda$, i.e. $X\sim\mathcal{P}(\lambda)$, if $$p(x)=e^{-\lambda}\dfrac{\lambda^x}{x!}$$ for $x \in S_X=\{0,1,2,\ldots\}$. **Poisson Distribution Properties:** - $\displaystyle\sum_{x\in S_X} p(x) = 1$ - ==Mean==: $\mathbb{E}(X) = \lambda$ - ==Variance==: $\mathbb{V}\operatorname{ar}(X)=\lambda$ > [!EXAMPLE] Example: Poisson Distribution > Over a 10-minute period, a counter records an average of 1.3 gamma particles per millisecond coming from a radioactive substance. To a good approximation, the distribution of the count, X , of gamma particles during the next millisecond is Poisson distributed. > *** > 1. Determine $\lambda$. > $$\begin{align*} > &\text{Let }X\text{ be the number of particles in the next millisecond}\\ > &\lambda=\mathbb{E}(X)=1.3 > \end{align*}$$ > > 2. Determine the probability of observing one or more gamma particles during the next millisecond. > $$\begin{align*} > P(X\geq 1)&=1-P(X=0)\\ > &=1-\frac{1.3^0}{0!}e^{-1.3}\\ > &=1-e^{-1.3}\\ > &\approx0.73 > \end{align*}$$ > > 3. Determine the variance of this number. > $$\begin{align*} > \mathbb{V}\operatorname{ar}(X)=\lambda=1.3 > \end{align*}$$ > > [!EXAMPLE] Example 2: Poisson Distribution > Suppose that the number of kangaroos killed per week on a particular stretch of road has Poisson distribution with parameter $\lambda=20$. > *** > 1. In the long-run, in what proportion of weeks will the number of dead roos be at most 10? > $$\begin{align*} > &\text{Let }X\text{ be the number of roadkills per week}\\ > &X\sim P(20)\\\\ > > P(X\leq10)&=\sum_{x=0}^{10}\frac{\lambda^x}{x!}e^{-\lambda}\\ > &=\frac{20^0}{0!}e^{-20}+\frac{20^1}{1!}e^{-20}+\cdots > \end{align*}$$ > > ```matlab > poisscdf(10,20) > ``` > $\approx0.011$ > > 2. In the long-run, in what proportion of weeks will the number of dead roos exceed 20? > $$\begin{align*} > P(X\gt20)=1-P(Y\leq20) > \end{align*}$$ > > ```matlab > 1-poisscdf(20,20) > ``` > $\approx0.44$ > > 3. In the long-run, in what proportion of weeks will the number of dead roos be between 10 and 20, inclusive? > $$\begin{align*} > &P(10\leq X\leq 20)\\ > &=P(X\leq20)-P(X\lt9) > \end{align*}$$ > > ```matlab > poisscdf(20,20)-poisscdf(9,20) > ``` > $\approx0.554$ > > 4. In the long-run, in what proportion of weeks will the number of dead roos be strictly between 10 and 20? > $$\begin{align*} > &P(10\lt X\lt20)\\ > &=P(X\leq19)-P(X\leq10) > \end{align*}$$ > > ```matlab > poisscdf(19,20)-poisscdf(10,20) > ``` > $\approx0.46$ #### Poisson Approximation to Binomial Since it was derived as a limit case of the Binomial distribution when $n$ is 'large' and $\pi$ is 'small', one can expect the Poisson distribution to be a good approximation to $\text{Bin}(n,\pi)$ in that case. ### 3.3 The Uniform Distribution A random variable is said to be **uniformly distributed** over an interval $[\alpha,\beta]$, i.e. $X \sim U_{[\alpha,\beta]}$ if its probability density function is given by $$f(x)=\left\{\begin{matrix} \frac{1}{\beta-\alpha} & \text{if } x \in [\alpha,\beta] \\ 0 & \text{otherwise} \end{matrix} \right. $$ The cdf for a uniform distribution is given by $$F(x)= \left\{\begin{array}{l l} 0 & \text{if } x < \alpha \\ \frac{x-\alpha}{\beta-\alpha} & \text{if }\alpha \leq x \leq \beta \\ 1 & \text{if } x>\beta \end{array} \right. $$ If $X \sim U_{[\alpha,\beta]}$: - $\mathbb{E}(X)= \dfrac{\alpha+\beta}{2}$ - $\mathbb{V}\operatorname{ar}(X)= \dfrac{(\beta-\alpha)^2}{12}$ **Uniform Distribution Properties:** The probability that $X$ lies in any subinterval $[a,b]$ of $[\alpha,\beta]$ is $\mathbb{P}(a< X< b)=\dfrac{b-a}{\beta-\alpha}$. If $X\sim U_{[a,b]}$, then **any linear transformation** of $X$ is also uniform. That is, if $Y=c+dX$ (for $d\neq0$), then $Y\sim U_{[c+ad,c+bd]}$. ![[Screenshot 2025-06-19 at 12.40.36 am.png|center|250]] > [!EXAMPLE] Example: Uniform Distribution > Buses arrive at a specified stop at 15-minute intervals starting at 7 A.M That is, they arrive at 7, 7:15, 7:30, 7:45, etc. > > If a passenger arrives at the stop at a time uniformly distributed between 7 and 7:30, find the probability that he waits less than 5 minutes for a bus. > > Let $X$ be minutes since 7am that passenger arrives at. > $X\sim U_{[0,30]}$ > > $$\begin{align*} > P(\text{wait less than 5min})&=P(10\leq X\leq15\text{ or }25\leq X\leq 30)\\ > &=\frac{5+5}{30}=\frac{1}{3} > \end{align*}$$ ### 3.4 The Exponential Distribution A random variable is said to be an **Exponential random variable** with parameter $\mu(\mu>0)$, i.e. $X\sim\text{Exp}(\mu)$, if its probability density function is given by $$f(x)=\left\{\begin{matrix} \frac{1}{\mu} e^{-\frac{x}{\mu}} & \text{if } x \geq 0 \\ 0 & \text{otherwise} \end{matrix} \right.$$ The cdf of an exponential distribution is $$F(x)= \left\{\begin{matrix} 0 & \text{if } x < 0 \\ 1-e^{-\frac{x}{\mu}} & \text{if } x \geq 0 \end{matrix} \right. $$ If $X \sim \text{Exp}(\mu)$: - $\mathbb{E}(X)= \mu$ - $\mathbb{V}\operatorname{ar}(X)= \mu^2$ > [!EXAMPLE] Example: Exponential Distribution > Suppose that, on average, 3 trucks arrive per hour to be unloaded at a warehouse. > *** > Let $X$ be time in minutes until next truck arrives. > $X\sim\text{Exp}(20)$ >
> 1. What is the probability that the time between the arrivals of two successive trucks will be less than 5 minutes? > $$\begin{align*} > P(X\lt5)&=\int^5_0\frac{1}{20}e^{-\frac{x}{20}}dx\\ > &=\left[-e^{-\frac{x}{20}}\right]^5_0\\ > &=-e^{-\frac{5}{20}}+1\\ > &\approx0.221 > \end{align*}$$ > > 2. What is the probability that the time between the arrivals of two successive trucks will be at least 45 minutes? > $$\begin{align*} > P(X\gt45)&=int_{45}^{\infty}\frac{1}{20}e^{-\frac{x}{20}}dx\\ > &=\left[-e^{-\frac{x}{20}}\right]^{\infty}_{45}\\ > &=0+e^{-\frac{45}{20}}\\ > &\approx0.105 > \end{align*}$$ ### 3.5 The Normal Distribution A random variable is said to be **normally distributed** with parameters $\mu$ and $\sigma\ (σ>0σ>0)$, i.e. $X \sim \mathcal{N}(\mu,\sigma)$ if its probability density function is given by $$f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ If $X \sim \mathcal{N}(\mu,\sigma)$: - $\mathbb{E}(X) = \mu$ - $\mathbb{V}\operatorname{ar} = \sigma^2$ The **Standard Normal distribution** is the Normal distribution with $\mu=0$ and $\sigma=1$. - Bell shaped ![[Screenshot 2025-06-19 at 1.12.08 am.png|center|200]] > [!EXAMPLE] Example: Normal Distribution > Suppose that $Z \sim \mathcal{N}(0,1)$. > > 1. What is $\mathbb{P}(Z \leq 1.25)$? > ```matlab > normcdf(1.25) > ``` > $=0.894$ >
> 2. What is $\mathbb{P}(Z > 1.25)$? > $=1-0.894$ > $=0.106$ >
> 3. What is $\mathbb{P}(-0.38 \leq Z < 1.25)$? > $P(Z<1.25)-P(Z<-0.38)$ > ```matlab > normcdf(1.25)-normcdf(-0.38) > ``` > $=0.542$ #### 68-95-99 Rule If $X \sim \mathbb{N}(\mu,\sigma)$, then $Z=\dfrac{X-\mu}{\sigma} \sim \mathcal{N}(0,1)$. The **68-95-99 rule** states that $\mathbb{P}(-1 < Z < 1) \simeq 0.68$, $\mathbb{P}(-2 < Z < 2) \simeq 0.95$ and $\mathbb{P}(-3 < Z < 3) \simeq 0.997$. **Standard Normal:** $$Z=\frac{x-\mu}{\sigma}$$ > [!EXAMPLE] Example: Standardisation of Normal > The time it takes a driver to react to the brake light on a decelerating vehicle follows a Normal distribution having parameters $\mu=1.25$ sec and $\sigma=0.46$ sec. > > In the long run, what proportion of reaction times will be between 1 and 1.75 sec? > *** > Let $X$ be reaction time $X\sim N(1.25, 0.46)$ > $$\begin{align*} > P(1 &=P(-0.544 &=P(Z<1.087)-P(Z\leq-0.544)\\ > &\approx0.568 > \end{align*}$$ > > ```matlab > z1=(1-1.25)/0.46 > z2=(1.75-1.25)/0.46 > normcdf(z2)-normcdf(z1) > ``` > **OR ALTERNATIVELY** > ```matlab > normcdf(1.75,1.25,0.46)-normcdf(1,1.25,0.46) > ``` #### Normal Quantiles The $\alpha$-**level quantile** of a standard normal distribution is the value $z_\alpha$ such that $\mathbb{P}(Z < z_\alpha) = \alpha$. ![[Screenshot 2025-06-19 at 1.35.11 am.png|center|200]] What is $z$ such that $P(Z\leq Z)=0.95$? ```matlab norminv(0.95) =1.6449 ``` > [!EXAMPLE] Example: Normal Quantiles > Assume again that the time it takes a driver to react to the brake light on a decelerating vehicle follows a Normal distribution having parameters $\mu=1.25$ sec and $\sigma=0.46$ sec. > > How long must reaction time be if it is to fall among the slowest $10\%$ of reaction times? > *** > Let $X$ be reaction time. $X\sim N(1.25,0.46)$. > Find $X$ such that $P(X\gt x)=0.1$. (i.e. $P(X\leq x)=0.9$). > > Using `norminv(0.9)`, z-score = $1.28$. > > Now, $P(Z\leq1.28)=0.9$, so, $P(\frac{x-\mu}{\sigma}\leq1.28)=0.9$. > $$\begin{align*} > \frac{x-1.25}{0.46}&=1.28\\ > x&=1.25+1.28+0.46\\ > &\approx1.84 > \end{align*}$$ > > **OR ALTERNATIVELY** > ```matlab > norminv(0.9,1.25,0.46) > ``` **Further Properties of Normal Distribution** Common quantiles: - $\mathbb{P} (Z > 1.645) = 0.05$ - $\mathbb{P} (Z > 1.96) = 0.025$ - $\mathbb{P} (Z > 2.575) = 0.005$ ![[Screenshot 2025-06-19 at 1.45.53 am.png|center|400]] Suppose $X_1 \sim \mathcal{N}(\mu_1,\sigma_1),\ X_2 \sim \mathcal{N}(\mu_2,\sigma_2)$ and $X_1$ and $X_2$ are **independent**. Then, for any real values $a$ and $b$, $aX_1+bX_2 \sim \mathcal{N}\left(a\mu_1+b\mu_2,\sqrt{a^2\sigma_1^2+b^2 \sigma_2^2}\right)$. > [!EXAMPLE] Example: Sum of Normal Variables > Suppose $\mu_1=40$ min, $\mu_2=50$ min, $\mu_3=60$ min, and $\sigma_1^2=10$ min$^2$, $\sigma_2^2=12$ min$^2$ and $\sigma_3^2=14$ min$^2$. > > What is the probability that the full task would take less than 160 min? > *** > $X_1+X_2+X_3\sim N(40+50+60,\sqrt{10+12+14})$ > $$\begin{align*} > P(X_1+X_2+X_3\lt160)&=P(Z<\frac{160-150}{6})\\ > &=P(Z<\frac{10}{6})\\ > &\approx0.953 > \end{align*}$$ ### 3.6 Checking if Data are Normally Distributed - A ==normal quantile plot== is the best way to check how close your data are to a normal distribution - This plot ==compares the data==, ordered from smallest to largest, to relevant quantiles from the standard normal distribution - If the data are close to normally distributed, they should be ==close to a straight line== on the normal quantile plot ![[Screenshot 2025-06-19 at 1.55.26 am.png|center|250]] ```matlab n=50; x=normrnd(0,1,n,1) i=((1:n)-0.5)/n norminv(i) plot(sort(x), norminv(i), '.') ``` > [!INFO] Check if the normal quantile plot is close to a straight line. > [!EXAMPLE] Example: Normal Quantile Plot > The data is clearly right-skewed. > > ![[Screenshot 2025-06-19 at 2.01.52 am.png|300]] #### Transforming Observations If our data are not normal, we can ==transform== it so there is less right-skew and the data are closer to being normal. > [!INFO] The Log Transformation > The log transformation is a good transformation to use. > *(calculate the log of all the values)* > > The log transformation takes the large range and squishes it together, making it smaller, changing the shape of the data. 1b:["$","div",null,{"className":"min-h-screen h-full w-full","children":[["$","div",null,{"className":"px-8 py-16 bg-lined border-b bg-background","children":["$","div",null,{"className":"flex flex-col md:flex-row md:items-center justify-between space-y-4","children":[["$","div",null,{"className":"flex flex-col space-y-2 md:mb-0","children":[["$","$L4",null,{"href":"/notes/MATH2859","className":"flex items-center gap-1 uppercase text-muted-foreground text-sm tracking-wide font-medium hover:underline underline-offset-4 w-fit","children":[["$","$L17",null,{"ref":"$undefined","iconNode":[["path",{"d":"m12 19-7-7 7-7","key":"1l729n"}],["path",{"d":"M19 12H5","key":"x3x0zl"}]],"className":"lucide-arrow-left size-4"}],"All ","MATH2859"," Notes"]}],["$","div",null,{"className":"flex gap-2 flex-col","children":[["$","h1",null,{"className":"font-semibold text-4xl flex items-center gap-1.5","children":[["$","$L17",null,{"ref":"$undefined","iconNode":[["path",{"d":"M6 22a2 2 0 0 1-2-2V4a2 2 0 0 1 2-2h8a2.4 2.4 0 0 1 1.704.706l3.588 3.588A2.4 2.4 0 0 1 20 8v12a2 2 0 0 1-2 2z","key":"1oefj6"}],["path",{"d":"M14 2v5a1 1 0 0 0 1 1h5","key":"wfsgrz"}],["path",{"d":"M10 9H8","key":"b1mrlr"}],["path",{"d":"M16 13H8","key":"t4e002"}],["path",{"d":"M16 17H8","key":"z1uh3a"}]],"className":"lucide-file-text size-8 shrink-0"}]," ","MATH2859"]}],["$","h2",null,{"className":"capitalize","children":["week 3"," •"," ","UNSW 25T2"]}]]}]]}],["$","$L23",null,{"categories":[],"courseId":"MATH2859","weekId":"week-3","resources":[{"title":"Week 0","file":"week-0"},{"title":"Week 1","file":"week-1"},{"title":"Week 2","file":"week-2"},{"title":"Week 3","file":"week-3"},{"title":"Week 4","file":"week-4"},{"title":"Week 5","file":"week-5"},{"title":"Week 7","file":"week-7"},{"title":"Week 8","file":"week-8"},{"title":"Week 9","file":"week-9"},{"title":"Week 10","file":"week-10"}]}]]}]}],["$","div",null,{"className":"px-8 py-4 bg-secondary/20 border-b","children":["$","p",null,{"className":"text-muted-foreground text-xs leading-5","children":[["$","span",null,{"className":"font-semibold","children":"Disclaimer:"}]," Images have been omitted to respect copyright. 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MATH2859

week 3 • UNSW 25T2