3. Continuous Random Variables


 Andra Sparks
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1 Statistics ad probability: Cotiuous Radom Variables A cotiuous radom variable is a radom variable which ca take values measured o a cotiuous scale e.g. weights, stregths, times or legths. For ay predetermied value x, P( X = x) = 0, sice if we measured X accurately eough, we are ever goig to hit the value x exactly. However the probability of some regio of values ear x ca be ozero. Probability desity fuctio (pdf): P(1.5 < X < 0.7 ) Probability of X i the rage a to b. Sice has to have some value Ad sice, For a pdf, for all. Cumulative distributio fuctio (cdf) : This is the probability of. Mea ad variace Expected value (mea) of : Variace of : Note that the mea ad variace may ot be well defied for distributios with broad tails. The mode is the value of where is maximum (which may ot be uique). The media is give by the value of x where.
2 Statistics ad probability: 3 Uiform distributio The cotiuous radom variable has the Uiform distributio betwee ad, with if f(x) { 1 x, for short. Roughly speakig,, if X ca oly take values betwee ad, ad ay value of withi these values is as likely as ay other value. Mea ad variace: for, ad Proof: Let y be the distace from the midpoit,. The sice meas add, ad the width be Usurprisigly the mea is the midpoit. * + ( ) Occurrece of the Uiform distributio 1) Waitig times from radom arrival time util a regular evet (see below) ) Egieerig toleraces: e.g. if a diameter is quoted "0.1mm", it sometimes assumed (probably icorrectly) that the error has a U(0.1, 0.1) distributio. 3) Simulatio: programmig laguages ofte have a stadard routie for simulatig the U(0, 1) distributio. This ca be used to simulate other probability distributios.
3 Statistics ad probability: 33 Example: Disk wait times I a hard disk drive, the disk rotates at 700rpm. The wait time is defied as the time betwee the read/write head movig ito positio ad the begiig of the required iformatio appearig uder the head. (a) Fid the distributio of the wait time. (b) Fid the mea ad stadard deviatio of the wait time. (c) Bootig a computer requires that 000 pieces of iformatio are read from radom positios. What is the total expected cotributio of the wait time to the boot time, ad rms deviatio? Solutio Rotatio time = 8.33ms. Wait time ca be aythig betwee 0 ad 8.33ms ad each time i this rage is as likely as ay other time. Therefore, distributio of the wait time is U(0, 8.33ms) (i.. 1 = 0 ad = 8.33ms). ; For 000 reads the mea time is ms = 8.3s. The variace is ms = 0.01s, so =0.11s. Expoetial distributio The cotiuous radom variable has the Expoetial distributio, parameter if: { Relatio to Poisso distributio: If a Poisso process has costat rate, the mea after a time is. The probability of ooccurreces i this time is
4 Statistics ad probability: 34 If is the pdf for the first occurrece, the the probability of o occurreces is also give by So equatig the two ways of calculatig the probability we have Now we ca differetiate with respect to givig hece :.the time util the first occurrece (ad betwee subsequet occurreces) has the Expoetial distributio, parameter. Occurrece 1) Time util the failure of a part. ) Times betwee radomly happeig evets Mea ad variace [ ] [ ] [ ] Example: Reliability The time till failure of a electroic compoet has a Expoetial distributio ad it is kow that 10% of compoets have failed by 1000 hours. (a) What is the probability that a compoet is still workig after 5000 hours? (b) Fid the mea ad stadard deviatio of the time till failure. Solutio (a) Let Y = time till failure i hours;
5 Statistics ad probability: 35 [ ] [ ] (b) Mea = = 9491 hours. Stadard deviatio = = = 9491 hours. Normal distributio The cotiuous radom variable has the Normal distributio if the pdf is: The parameter is the mea ad ad the variace is. The distributio is also sometimes called a Gaussia distributio. The pdf is symmetric about. X lies betwee ad with probability 0.95 i.e. X lies withi stadard deviatios of the mea approximately 95% of the time. Normalizatio [oexamiable] caot be itegrated aalytically for geeral rages, but the full rage ca be itegated as follows. Defie I ( x) dx e dx e x The switchig to polar coordiates we have
6 Statistics ad probability: 36 I dx e e x r 0 dy e y dxdy e x y 0 0 rdrd e r 0 rdre r Hece I = ad the ormal distributio itegrates to oe. Mea ad variace The mea is because the distributio is symmetric about (or you ca check explicitly by itegratig by parts). The variace ca be also be checked by itegratig by parts: [ ] Occurrece of the Normal distributio 1) Quite a few variables, e.g. huma height, measuremet errors, detector oise. (Bellshaped histogram). ) Sample meas ad totals  see below, Cetral Limit Theorem. 3) Approximatio to several other distributios  see below. Chage of variable The probability for X i a rage aroud is for a distributio is give by The probability should be the same if it is writte i terms of aother variable. Hece
7 Statistics ad probability: 37 Stadard Normal distributio There is o simple formula for, so umerical itegratio (or tables) must be used. The followig result meas that it is oly ecessary to have tables for oe value of ad. If, the This follows whe chagig variables sice hece Z is the stadardised value of X; N(0, 1) is the stadard Normal distributio. The Normal tables give values of Q=P(Z z), also called (z), for z betwee 0 ad Outside of exams this is probably best evaluated usig a computer package (e.g. Maple, Mathematica, Matlab, Excel); for historical reasos you still have to use tables. Example: Usig stadard Normal tables (o course web page ad i exams) If Z ~ N(0, 1): (a) (b) = = = (by symmetry) (c) = (d) = (1.5)  (0.5) = =
8 Statistics ad probability: 38 (e)  Usig iterpolatio: ( ) (f) Usig tables "i reverse",. (g) Fidig a rage of values withi which lies with probability 0.95: The aswer is ot uique; but suppose we wat a iterval which is symmetric about zero i.e. betwee d ad d. Tail area = 0.05 P(Z d) = (d) = Usig the tables "i reverse", d = rage is to P=0.05 P=0.05 Example: Maufacturig variability The outside diameter, X mm, of a copper pipe is N(15.00, 0.0 ) ad the fittigs for joiig the pipe have iside diameter Y mm, where Y ~ N(15.07, 0.0 ). (i) Fid the probability that X exceeds mm. (ii) Withi what rage will X lie with probability 0.95? (iii) Fid the probability that a radomly chose pipe fits ito a radomly chose fittig (i.e. X < Y). Solutio (i) ( ) (ii) From previous example lies i (1.96, 1.96) with probability i.e. ( )
9 Statistics ad probability: 39 i.e. the required rage is 14.96mm to 15.04mm. (iii) For we wat ). To aswer this we eed to kow the distributio of Distributio of the sum of Normal variates Remember tha meas ad variaces of idepedet radom variables just add. So if are idepedet ad each have a ormal distributio, we ca easily calculate the mea ad variace of the sum. A special property of the Normal distributio is that the distributio of the sum of Normal variates is also a Normal distributio. So if are costats the: c X c X c X ~ N( c c, c c c ) Proof that the distributio of the sum is Normal is beyod scope. Useful special cases for two variables are If all the X's have the same distributio i.e. 1 = =... = =, say ad 1 = =... = =, say, the: (iii) All c i = 1: X 1 + X X ~ N(, ) (iv) All c i = 1/: X = X X X 1 ~ N(, /) The last result tells you that if you average idetical idepedet oisy measuremets, the error decreases by. (variace goes dow as ). Example: Maufacturig variability (iii) Fid the probability that a radomly chose pipe fits ito a radomly chose fittig (i.e. X < Y). Usig the above results Hece ( )
10 Statistics ad probability: 310 Example: detector oise A detector o a satellite ca measure T+g, the temperature T of a source with a radom oise g, where g ~ N(0, 1K ). How may detectors with idepedet oise would you eed to measure T to a rms error of 0.1K? Aswer: We ca estimate the temperature from detectors by calculatig the mea from each. The variace of the mea will be 1K / where is the umber of detectors. A rms error of 0.1K correspods to a variace of 0.01 K, hece we eed =100 detectors. Normal approximatios Cetral Limit Theorem: If X 1, X,... are idepedet radom variables with the same distributio, which has mea ad variace (both fiite), the the sum X i i1 teds to the distributio as. Hece: The sample mea X = for large. 1 X i i 1 is distributed approximately as N(, /) For the approximatio to be good, has to be bigger tha 30 or more for skewed distributios, but ca be quite small for simple symmetric distributios. The approximatio teds to have much better fractioal accuracy ear the peak tha i the tails: do t rely o the approximatio to estimate the probability of very rare evets. Example: Average of samples from a uiform distributio:
11 Statistics ad probability: 311 Normal approximatio to the Biomial If X ~ B(, p) ad is large ad p is ot too ear 0 or 1, the X is approximately N(p, p(1p)). p p The probability of gettig from the Biomial distributio ca be approximated as the probability uder a Normal distributio for gettig i the rage from to. For example ca be approximated as where is the Normal distributio: Example: I toss a coi 1000 times, what is the probability that I get more tha 550 heads? Aswer: The umber of heads has a biomial distributio with mea p=500 ad variace So the umber of heads ca be approximated as. Hece ( )
12 Statistics ad probability: 31 Quality cotrol example: The maufacturig of computer chips produces 10% defective chips. 00 chips are radomly selected from a large productio batch. What is the probability that fewer tha 15 are defective? Aswer: the mea is, variace. So if is the umber of defective chips, approximately, hece ( ) [ ] This compares to the exact Biomial aswer. The Biomial aswer is easy to calculate o a computer, but the Normal approximatio is much easier if you have to do it by had. The Normal approximatio is about right, but ot accurate. Normal approximatio to the Poisso If Poisso parameter ad is large (> 7, say), the has approximately a distributio.
13 Statistics ad probability: 313 Example: Stock Cotrol At a give hospital, patiets with a particular virus arrive at a average rate of oce every five days. Pills to treat the virus (oe per patiet) have to be ordered every 100 days. You are curretly out of pills; how may should you order if the probability of ruig out is to be less tha 0.005? Solutio Assume the patiets arrive idepedetly, so this is a Poisso process, with rate 0. / day. Therefore, Y, umber of pills eeded i 100 days, ~ Poisso, = 100 x 0. = 0. We wat, or ( ) uder the Normal approximatio, where a probability of correspods (from tables) to.575. Sice this correspods to., so we eed to order pills. Commet Let s say the virus is deadly, so we wat to make sure the probability is less tha 1 i a millio, A ormal approximatio would give 4.7 above the mea, so pills. But surely gettig just a bit above twice the average umber of cases is ot that ulikely?? Yes ideed, the assumptio of idepedece is extremely ulikely to be valid. Viruses ted to be ifectious, so occurreces are defiitely ot idepedet. There is likely to be a small but sigificat probability of a large umber of people beig ifected simultaeously a much larger umber of pills eeds to be stocked to be safe. Do t use approximatios that are too simple if their failure might be importat! Rare evets i particular are ofte a lot more likely tha predicted by (too) simple approximatios for the probability distributio.
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