1 Density functions, cummulative density functions, measures of central tendency, and measures of dispersion


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1 Density functions, cummulative density functions, measures of central tendency, and measures of dispersion densityfunctionsintro.tex October, 9 Note tat tis section of notes is limitied to te consideration of a single and continuous random variable. Remember from earlier tat te variable X is a onedimensional, continuous random variable if tere exists a function f(x) suc tat f(x) 8 x in te interval x, and te probability tat (a x b) is Pr(a x b) = Z b a f(x)dx Te function f(x) is called a density function (or a probability density function).
2 . Examples of density functions te normal: f(x) = p e (x u) = is a well known density function were and u are parameters in te density function. Wo rst expressed tis formula? Wo rst used it as a density function?
3 Graps of te normal probability density functions are te familiar bellsaped curves. Te following plots sow te probabilitydensity functions NormalDen (x; ; ), cumulative distribution functions NormalDist (x; ; ) for te parameters (; ) = (; ) ; (; 5) ; (; :5) ; (; ) x Normal density functions 4 4 x Normal distribution functions Z t But, be warned tat p e (t u) = dt does not ave a closedform solution. Lets start more simply. Note tat p = p, ones sees te density function written bot ways, and te distinction is easy to miss. 3
4 .. Consider te Following Four Density Functions Example Assume and g(x) = if x < or x > 3 So, g(x) is not a density function, but Z 3 g(x) = 3x; x 3 3xdx = :5x j 3 = :5(9) = 3:5 6= is. f(x) = 3x 3:5 and Tat is f(x) = if x < or x > 3 f(x) = :x if x 3 4
5 Note te importance of being explicit about were f(x) =. Example f(x) = (3 + x) x 4 8 f(x) = oterwise Is tis a density function? f(x) 8 x Integrate it to see if te area under it is one. Z f(x)dx = So, yes. Z = Z 4 8 = Z 4 dx + Z (3 + x) dx + dx 8 4 (3 + x) dx = 8 3x + x j 4 = [( + 6) (6 + 4)] 8 5
6 Example 3 f(x) = se s(x n) exp e s(x n)i s > Te parameters are s and n. Use Matematica to grap tis function and see sow it canges as s and n cange. Z f(x) 8 x Z f(x)dx = = exp se s(x e s(x n) exp n)i e s(x n)i dx Wy? Recollect tat Z m (x)e m(x) dx = e m(x) So Z f(x)dx = lim exp b!+ = + = e s(b n)i lim exp a! e s(a n)i So f(x) = se s(x n) exp e s(x n)i is a density function. It is called te Extreme Value Distribution. It is te foundation of logit models of coice. 6
7 Here are tree examples of te Extreme Value Distribution. All ave n =, te blue is s =, te red is s = :5 and te black is s =. y x Here are tree more examples. All ave s =, te red is n = is n = 3 and te black is n =., te blue y x 7
8 Example 4 f(x) = + (x u) < x < f(x) 8 x Z Z f(x)dx = + (x u) dx = Note arctan ( )= tan ( ). lim arctan (b u) b! lim arctan (a u) a! Z f(x)dx = = + = f(x) = +(x u) is called te Caucy distribution. CaucyLourentz Distribution. from ttp://en.wikipedia.org/wiki/augustin_caucy, October, 7 Augustin Louis Caucy Notice te great combover. Tis is actually a restricted Caucy. Incorporating a scale parameter > te Caucy probability density function is f(x) = x +! 8
9 Born August 789, Dijon, France Died, 3 May 857 (aged 67) Now you migt ask wo Lourentz is, ttp://en.wikipedia.org/wiki/hendrik_lorentz, October, 7 Hendrik Antoon Lorentz Born July 8, 853, Arnem, Neterlands Died February 4, 98 (aged 74) Nobel Prize for Pysics (9) Te median of tis distribution is. Te Caucy probability density function is symmetric about and as a unique maximum at u. If u = it simpli es to te Willy/Marsall distribution. (Willy and Marsal were two former students.) f(x) = + x Te Caucy distribution is kind of cool: variance. 3 it does not ave a mean or a. 3 Or, for tat matter, any moments. We ave yet to talk about moments. 9
10 .3 Given te density function, f(x),.4.5 Pr(a x b) = Z b f(x)dx. so For example f(x) = :x x 3 = oterwise a Pr( x ) = Z (:x) dx = :x j = : [4 ] = :333 = 3 For example f(x) = (3 + x) 8 x 4 = oterwise so Pr( < x < 3) = Pr ob( x 3) = = Z 3 8 Z 3 f(x)dx (3 + x) dx = 8 3x + x j 3 = [(9 + 9) (6 + 4)] 8 = 8 8 = 8 8
11 For example 3 f(x) = se s(x n) exp s > e s(x n)i so Pr ( < x < b) = Pr (x b) = Z b = exp f(x)dx = Z b se s(x n) exp e s(b n)i lim exp a! e s(x e s(a n)i dx n)i but because as a! ; Terefore, as a! ; So, as a! ; and lim exp a! exp e s(a n)i = s (a n)! e s(a n)! e s(a n)! e s(a n)i! Terefore, for te Extreme Value Distribution Pr(x b) = exp e s(b n)i
12 For example, Pr(x n) = exp e s(n n)i = exp e = exp [ ] = e = e Consider n+ :3665 s x n + :3665 s? (were did tis come from?) Wat is te probability tat Pr ob x n + :3665 s = exp = exp e e s(n+ :3665 s n) i s( :3665 s ) i = exp e :3665 = :5 Wat is n + :3665 s? It is te median of te EV distribution. How did I gure out tat tis is te median of te EV distribution? Since Pr(x b) = exp e s(b n)i at median Pr(x median) = exp e s(median n)i = :5 by de nition of te median. Simplify by letting = s (median n)! exp e = :5 Solve for by taking te ln of eac side! = :36659 but = s (median n) : Solve for median = n + :36659 s
13 For example 4 f(x) = ( + x ) u = so Z Z Pr(x > ) = f(x)dx = = lim arctan b b! = + x dx [arctan ] = = In words, zero is te median of te Willy/Marsall distribution..6 3
14 .7 Given te density function f(x), wat is te probability tat X is less tan or equal to x, were x is a speci c value of X? Denote tis probability were Pr(X x) = F (x) < x < So te probability tat x b is Pr(X x) = F (x) = xz f(t)dt Pr(X b) = F (b) = Z b f(t)dt = Z b f(x)dx F (x) is called te cumulative density function for x. We ave already calculated te CDF (cumulative density function) for te Extreme Value Distribution and determined tat F (x) = exp e s(x n)i 4
15 Wat is te CDF for? For Example f(x) = :x x 3 = oterwise so xz F (x) = f(t)dt F (x) = if x xz F (x) = :t dt = :t j x = :x if < x < 3 F (x) = if x 3 5
16 For Example f(x) = (3 + x) 8 x 4 = oterwise so F (x) = if x < F (x) = if x 4 F (x) = = = xz (3 + t) dt = 3t + t j x 8 8 3x + x (6 + 4) 8 3x + x if x 4 8 6
17 For Example 3 We ave already determined tat for te Extreme Value Distribution F (x) = exp e s(x n)i 7
18 For Example 4 Trig Looking it up: Te CDF for te Caucy is F (x) = arctan(x ) + 8
19 For te normal distribution Tere is also no closedform solution for F (x) for te normal distribution. Tat is xz f(t)dt does not ave a closedform solution if f(x) = p e (x u) = xr However, given speci c values for u and, one can numerically solve f(t)dt for any x. Most statistics books provide tables for x given u = and = (te standard normal). Now tere are web sites tat provide tese tables interactively, you plug in u; and x and out comes te probability tat X x..8 9
20 .9 Note tat if ten because F (a) = Pr(x < a) F (a) = Pr(x > a) F (a) + ( F (a)) =
21 . Measures of central tendency Measures of central tendency are ways to describe one aspect of a distribution, f(x). Tree measures of central tendency are: mean (expected value) median mode Te mean (expected value) of any continuous random variable x wit distribution f(x) is de ned as E(X) = Z xf(x)dx. Wat does E(X) mean (no pun intended)? If one randomly sampled one X, one would not expect it to be E(X), so it is not te number you "expect." But, if one randomly sampled N X s, one would expect te average value of te sampled x s to! E(X) as N!. Te median of a continuous random variable x wit density function f(x) is de ned as u were uz f(x)dx = Z = f(x)dx u Given my de nition of te median, Is te median always a unique number, or can it be a range? Sould we cange my de nition so tat it only exists wen is unique? If a density function as a unique global max, te value of x tat max f(x) is called te mode. Very loosely speaking, te mode is te most common value for x (remember tat if x is continuously distributed te probability of observing any speci c value of x is zero). If te distribution is symmetric, mean = median. For some distributions, mean = median = mode: e.g. te normal and te logistic. Not tat some distributions do not ave a mean.
22 Te mean of (example ) f(x) = :x x 3 = oterwise so E(X) = = Z Z Z 3 = + = xf(x)dx = xdx + Z 3 Z Z 3 xf(x)dx + Z x (:) xdx + xdx 3 Z xf(x)dx + xf(x)dx (:) x dx + = : 3 x3 j 3 = :74x 3 j 3 Te mean of (example ) f(x) = (3 + x) 8 x 4 = oterwise 3 is E(X) = = Z Z xf(x)dx = xdx + Z 4 Z Z 4 xf(x)dx + Z xf(x)dx + xf(x)dx Z x (3 + x) dx + xdx 8 = Z x + x dx + = 3 8 x x3 j 4 = 8 (6) + 3 (64) = [(4 + 4:666) 8 (6 + 5:333)] = 8 = 3: [66:666 :333] 3 (4) + 3 (8)
23 For te Extreme Value Distribution (example 3) f(x) = se s(x n) exp e s(x n)i so E(X) = Z xf(x)dx = Z xse s(x n) exp e s(x n)i dx = Z xm (x)e m(x) dx =? were m(x) = e s(x n) I know tat E(X) = n+ s were is te Euler constant (:577). But I ave been unable to derive it analytically. Can you? 3
24 Can you do it for a speci c values of n and s? For example, if n = and s =, E(X) = Z xf(x)dx = Z xe (x) exp e (x)i dx Given E(X) = n + s, if s = and n =, E(X) = n + s = = :577:::: Looking at te grap of e (x) exp e (x), one migt guesstimate tat E(X) = :577:::, is a bit greater tan zero. y x 4
25 Z R In tis special case (n = and s = ), E(X) = xf(x)dx = xe x exp [ e x ] dx, my mat software can gure out te correct answer. Note tat below I am taking te de nite integral over an every increasing range, and tat te answer is approacing. (My software will get mad if I plug in positive and negative in nity) R xe x exp [ e x ] dx = : 75 5 R xe x exp [ e x ] dx = :94 48 R 5 5 xe x exp [ e x ] dx = :536 9 R xe x exp [ e x ] dx = :576 7 R xe x exp [ e x ] dx = :577 If I wanted to generalize I would continue to assume s = and try to take te de nitie integral for di erent values of n to see ow tis a ects te answer. Hopefully, I would nd tat E(X) = n +, and, ten if tat works, I would start playing wit di erent values for s. As an aside, note tat te mode is n. 5
26 For te Caucy Distribution (example 4) It does not ave a mean. Wow. For a bit of explanation as to wy, see te asides at te end of tis lecture. 6
27 .. Medians We ave already determined te median for te Extreme Value Distribution (example 3) and te Willy/Marsall distribution (example 4). If u is te median Pr(x u) = :5 For te Extreme Value Distribution f(x) = se s(x n) exp e s(x n)i Earlier we determined tat Recollect tat median = n + : 3665 s mean = E(X) = n + s were is te Euler constant ~:5776: So, mean 6= median and mean median = n + :5776 s n + : 3665 = :7 s s 7
28 Earlier we determined tat te median for te Willy/Marsall distribution is. More generally te median of te Caucy distribution is  it as a median, just not a mean. Determine te mode for te four example distributions? 8
29 . Measures of dispersion One question we often ask about distributions is: How dispersed or spread out are te values of X? Te most common measure of spread is variance. Te variance is a measure of dispersion around te mean (E(X)). V ariance E (x E(X)) i expected value of (x E(X)) Anoter possible measure of te dispersion around te mean is E [jx E(X)j]. One could alternatively de ne measures of dispersion around te median or mode. 9
30 It can be sown tat if f(x) is te density function for x and g(x) is some function of x E [g(x)] = Z g(x)f(x)dx (tis is a formula we will use often.) 4 We used a special case of tis formula to get E(X) = mean. Tat is, if g(x) = x E [X] = Z xf(x)dx We can also use it to get E (x E(X)) i In tis case, g(x) = (x E(X)) and E (x E(X)) i = Z (x E(X)) f(x)dx 4 Investigate wy it is true. 3
31 Let s calculate te variance of x for our earlier examples. For example f(x) = :x x 3 = oterwise Te variance is Since E(X) =, = Z (x E(X)) f(x)dx = = = Z Z Z 3 = :5 (x ) f(x)dx Z 3 (x ) dx + (x ) (:x) dx (x ) (:x) dx + Z 3 (x ) dx 3
32 So, wat does tis mean? In words, wat does a variance of :5 mean? For example f(x) = (3 + x) 8 x 4 = oterwise Since E(X) = 3:74, te variance of x is = = = = Z Z Z Z 4 = :37846 (x E(X)) f(x)dx (x 3:74) f(x)dx 4Z (x 3:74) dx + (x 3:74) 8 (x 3:74) (3 + x) dx 8 Z (3 + x) dx + 4 (x 3:74) dx 3
33 For Example 3 Can we gure out te variance for te Extreme Value Distribution? f(x) = se s(x n) exp e s(x n)i Since E(X) = n + s, were is Euler s constant = = Z Z (x E(X)) se s(x n) exp x (n + s ) se s(x n) exp e s(x e s(x n)i dx n)i dx Luckily for us, someone (?) as already determined tat = 6s 33
34 Example 4 (te Caucy distribution). Since it does not ave an E[X], it cannot ave a wellde ned variance because te variance is a function of te E[X]. For te normal density function f(x) = p e (x u) = te variance is simply. 34
35 . Some asides.. Is tere a way to derive te E[X] directly from te CDF? Te answer is yes (from MGB), as tere must be: E[X] = + Z [ F (x)] dx Z F (x)dx Can you intuite tis? Try it out and see if it works. If I remember correctly, tis formula is useful if te one as a mixed distribution (discrete and continuous). Maybe, you could use it to get E[X] for te Extreme Value distribution, were F (x) = exp + e s(x n) Z Z + Z, E[X] = [ F (x)] dx F (x)dx = exp e s(x n) dx Z exp e s(x n) dx. NOPE 35
36 .. Some probabilities of interest For some distributions, one you determine Pr[(E[X] ) x (E[X] + )] were is de ned as te variance, or Pr[(E[X] ) x (E[X] + )]. Te rst is te probability tat a randomly drawn X lies between (E[X] ) and (E[X] + ) For a given f (x) ; tis is te probability tat...? At tis point, you sould ave te tools to gure tese out? Will te value of Pr[(E[X] ) x (E[X] + )] be a function of f(x)? or be invariant to te form of te density function? For example, in terms of te CDF, F (X) Pr[(E[X] ) x (E[X] + )] = F (E[X] + ) F ((E[X] )) 36
37 It looks like it is a function of F (X), so not invariant to te form of te distribution E[X] and variance as areas Recollect tat if f(x) is a density function ten is te area under te density function. + Z f(x)dx =, were + Z f(x)dx E(X) = xf(x)dx can be tougt of as an area, te area under te function xf(x). R And, variance is te area under (x can be viewed as areas. E[X]) f(x). Tat is, mean and variance So te mean will not be well de ned unless te area under xf(x) is well de ned, and te variance will not be well de ned unless te area under (x E[X]) f(x) is well de ned  tis will sometimes be an issue, mostly wit density functions were f(x) > < x < +. Consider te uniform distribution on te unit interval, a very simple example. For tis density, xf(x) is a straigt line from (; ) to (; ), and te area under it is :5 = E[X] 5 Many students tend to tink, incorrectly, tat Pr[(E[X] ) x (E[X]+)] is invariant to te form of F (X) 37
38 xf(x) te function xf(x) = x() x Note tat in regions were x < and f(x) >, xf(x) <, so if te xf(x) spans any of te negative region, 38
39 it will ave negative area in tose sections. 39
40 Consider, for example, te uniform distribution on te interval ( :5; +:5). For tis density function, xf(x) is a straigt line from ( :5; :5) to (:5; :5); te area under it to te left of zero is :5 and to te rigt of zero is +:5 and E[X] =. xf(x)=x x.5 Z:5 :5 xdx = : 4
41 Consider te standard normal f(x) = p e (x) = f(x) Te area under tis curve is  tat s wy it is a density function x For te standard normal xf(x) = xf(x) x p e (x) = x and te "area under tis is zero, Z ( x p e (x) = )dx =, so te correct answer 4
42 for te mean of te standard normal. How about te variance (wic we know is ). It is te area under (x ) f(x) = p x e (x) = y x Z 5 ( p x e (x) = )dx = :999 98, wow, te correct answer. 5 Furter note tat we can always write E(X) = = = Z xf(x)dx Z + Z xf(x)dx + xf(x)dx + Z Z xf(x)dx jxj f(x)dx 4
43 for te extreme value (n = and s = ), E(X) is te area under xe (x) exp e (x) xf(x)
44 As from before R 5 5 xe x exp [ e x ] dx = : Density functions were f(x) > < x < +..5 Te normal, te Caucy and te Extreme Value are all density function were every value of X as positive density. Imagine you are trying to create a new density function were f(x) > 8x. Te area under many functions tat meet te condition f(x) > 8x ave + Z f(x)dx = +, so cannot be density functions. One must coose carefully: one must coose a function so tat as one integrates over larger and larger ranges of X, te area under te function must forever increase, but at te same time te area can never be greater tan one  neat trick. Te normal, te Caucy and te Extreme value, since tey all are density functions, do te trick. But te fact tat te area under f(x) is well de ned, does not imply tat te areas under xf(x) and (x E[X]) f(x) are well de ned. 6 Tese areas are well de ned for te normal and te Extreme value, but tey are not for te Caucy  tat is wy te Caucy does not ave a mean. Put loosly, te reason te Caucy does not ave a mean is tat and + Z xf(x)dx = +  te tails of xf(x) are too fat. Z xf(x)dx = 6 Note tat if te area under xf(x) is not well de ned, ten (x E[X]) f(x) is not well de ned. 44
45 Te following is from Wikipedia. Go gure! 45
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