1, 0 r 1 f R (r) = 0, otherwise. 1 E(R 2 ) = r 2 f R (r)dr = r 2 3

Transcription

1 STAT We are given that a circle s radius R U(, ). the pdf of R is {, r f R (r), otherwise. The area of the circle is A πr. The mean of A is E(A) E(πR ) πe(r ). The second moment of R is ( ) r E(R ) r f R (r)dr r 3 dr 3 3. E(A) πe(r ) π 3. The variance of A is V (A) V (πr ) π V (R ). How do we calculate V (R )? Use the variance computing formula for the random variable R ; i.e., The fourth moment of R is V (R ) E(R 4 ) [E(R )] E(R 4 ) ( ). 3 E(R 4 ) r 4 f R (r)dr V (A) π V (R ) π [ 5 ( ) r r 4 5 dr 5 5. ( ) ] 4π Note: You could also get E(R ) and E(R 4 ) using the mgf of R U(, ); i.e., m R (t) et, t ; t i.e., calculate nd and 4th derivatives and evaluate at t. You should get the same answers for E(R ) and E(R 4 ) (a) The change in depth Y U(, ); i.e., θ and θ. the pdf of Y is f Y (y) 4, < y, otherwise. (b) The general expression for the cdf of Y is which we must calculate for all y R. Case : When y, f Y (t)dt f Y (t)dt, dt. PAGE

2 STAT 5 PDF /4 CDF y y Case : When < y <, f Y (t)dt dt + 4 dt + t 4 y y ( ) 4 y + 4. Case 3: When y, f Y (t)dt dt + t 4 dt + dt. Summarizing, the cdf of Y is, y y + 4, < y <, y. The pdf and cdf of Y are shown side by side in the figure above. Here is the R code I used: y seq(-,,.) pdf (/4)+y-y # Plot PDF plot(y,pdf,xlab"y",ylab"pdf",type"l",xaxt"n",yaxt"n",xlimc(-3,3),ylimc(,), cex.lab.5) abline(h) axis(,atc(-,,),labels c("-","","")) axis(,atc(,/4),labels c("","/4")) PAGE

3 STAT 5 # Plot CDF cdf (y+)/4 plot(y,cdf,xlab"y",ylab"cdf",type"l",xaxt"n",xlimc(-3,3),ylimc(,), cex.lab.5) abline(h) lines(c(,3),c(,),lty) axis(,atc(-,,),labels c("-","","")) Let Y denote the SAT mathematics exam score. We assume Y N (µ 48, ). In part (a), we want to calculate P (Y < 55) 55 e ( y 48 ) π() } {{ } N (48, ) pdf dy F Y (55) where F Y ( ) denotes the N (µ 48, ) cdf. In R, this is calculated as > pnorm(55,48,) [] about 76 percent of the students will score 55 or below. Here is the pdf of Y with P (Y < 55) shaded: PDF Here is the R code I used to make the figure above: y seq(3,83,.) pdf dnorm(y,48,) plot(y,pdf,type"l",xlab"y",ylab"pdf",xaxpc(8,78,6),cex.lab.5) abline(h) # Add shading corresponding to P(Y<55) x seq(3,55,.) y dnorm(x,48,) polygon(c(3,x,55),c(,y,),col"lightblue") points(x55,y,pch9,cex.5) y PAGE 3

4 STAT 5 (b) In part (a), we determined that 55 is the quantile of the N (µ 48, ) distribution; i.e., P (Y < 55) In part (b), we want to find this same quantile from the distribution of X N (µ 8, 6 ); i.e., the distribution of X ACT math score. From R, > qnorm( ,8,6) [] In this problem, we want to show the N (µ, ) pdf f Y (y) has points of inflection at y µ ±. We can do this by finding the second derivative of f Y (y). The first derivative of f Y (y) is f Y (y) d dy f Y (y) d dy [ e ( y µ ) ] π π d ] [e ( y µ ) dy π e ( y µ ) ( ) ( y µ ) ( ) To find the second derivative, use the product rule: f Y (y) d dy f Y (y) d [ (y µ) dy π 3 e ( y µ ) ] [ ] ( π 3 e ( y µ ) (y µ) + e ( y µ ) π 3 π 3 e ( y µ ) (y µ) + π 5 e ( y µ ) [ (y ) π 3 e ( y µ ) µ ]. (y µ) π 3 e ( y µ ). ) ( y µ ) ( ) Now, set f Y (y) equal to zero and solve for y; we get [ (y ) π 3 e ( y µ ) µ set ] this can never be ( ) y µ ( y µ ) y µ ±. Note that if (y µ)/ +, then If (y µ)/, then y µ y µ +. y µ y µ. This argument shows the points of inflection on the N (µ, ) pdf f Y (y) are at y µ ±. For example, the points of inflection on the N (µ 48, ) pdf in Exercise 4.77 are y 38 and y 58. PAGE 4

5 STAT The gamma function, which is defined for α >, is Γ(α) y α e y dy. In part (a), put in α ; we get Γ() y e y dy e y dy e y ( ) lim y e y ( ). In part (b), suppose α >. In the integral use integration by parts with Γ(α) y α e y dy, u y α du (α )y α dy dv e y v e y. With these selections, as claimed. Γ(α) y α e y dy y α e y + (α ) (α ) y (α ) e y dy } {{ } Γ(α ) y α e y dy (α )Γ(α ), 4.9. The time to complete the operation is a random variable Y exponential(β ). Recall that E(Y ) β and V (Y ) β. Define C + 4Y + 3Y. We want to find E(C) and V (C). Finding E(C) is easy. Note that E(C) E( + 4Y + 3Y ) + 4E(Y ) + 3E(Y ). We can get the second moment of Y quickly from the variance computing formula. Recall and V (Y ) E(Y ) [E(Y )] E(Y ) V (Y ) + [E(Y )]. E(Y ) + () E(C) + 4() + 3(). Getting V (C) is harder. One way is to use the variance computing formula for the function C; i.e., V (C) E(C ) [E(C)] E(C ) (). PAGE 5

6 STAT 5 We now have to get E(C ). We have E(C ) E[( + 4Y + 3Y ) ] E( + 6Y + 9Y 4 + 8Y + 6Y + 4Y 3 ) + 8E(Y ) + E(Y ) + 4E(Y 3 ) + 9E(Y 4 ). We know E(Y ) and E(Y ). We have to get the third and fourth moments E(Y 3 ) and E(Y 4 ), respectively. We could get these using the moment generating function, but it might be easier to just calculate the moments directly. Note that Also, E(Y 3 ) E(Y 4 ) y 3 e y/ dy y 4 e y/ dy y 4 e y/ dy Γ(4)4 6. y 5 e y/ dy Γ(5)5 4. E(C ) + 8() + () + 4(6) + 9(4) 43. Finally, V (C) E(C ) () 43 () 9. Note: Another way to get V (C) is to use the definition of variance for the random variable C. Recall that V (C) E[(C ) ] E[( + 4Y + 3Y ) ] You can do this integral numerically in R: ( + 4y + 3y ) e y/ dy. > integrand <- function(y){(+4*y+3*y^-)^*(/)*exp(-y/)} > integrate(integrand,lower,upperinf) 9 with absolute error < Let Y denote the one-hour CO concentration (measured in ppm). In part (a), we assume Y exponential(β 3.6) and want to calculate P (Y > 9). The easiest way is to note that P (Y > 9) P (Y 9) F Y (9), where F Y ( ) is the cdf of Y. Recall the cdf of Y exponential(β) is given by {, y e y/β, y >. ( P (Y > 9) F Y (9) e 9/3.6) e 9/ PAGE 6

7 STAT 5 If you didn t remember to use the cdf here, you could always just calculate P (Y > 9) directly using the pdf: P (Y > 9) 3.6 e y/3.6 dy ) ( 3.6e y/ You can also use R to check your work: > -pexp(9,/3.6) [].885 e y/ e 9/3.6 lim y e y/3.6 e 9/ In part (b), we perform the same calculation but with β.5 ppm instead: ( P (Y > 9) e 9/.5) e 9/ The support of Y is R {y : y > }, so this makes me think of the gamma family of distributions. Look at the kernel of the pdf: y e y y 3 e y/(/). This is the kernel of the gamma distribution with α 3 and β /. Sure enough, the constant out front in the gamma pdf Γ(3) ( ) 3 ( ) 4. 8 Y gamma(α 3, β /). We have E(Y ) αβ 3 and V (Y ) αβ Suppose Y gamma(α, β). The hardest part about this problem is keeping a and α separate (the authors should have used another symbol other than a here). In part (a), suppose α + a >. From the definition of mathematical expectation, as claimed. E(Y a ) y a Γ(α)β α yα e y/β dy gamma(α,β) pdf Γ(α)β α Γ(α)β α yα+a e y/β dy y α+a e y/β dy } {{ } Γ(a+α)β α+a Γ(α)β α Γ(a + α)βα+a βa Γ(α + a), Γ(α) PAGE 7

8 STAT 5 (b) When we say y α+a e y/β dy Γ(α + a)β α+a in part (a), we are using the fact that Γ(α + a)β α+a yα+a e y/β dy, that is, the gamma(α+a, β) pdf integrates to. Recall that the shape parameter in the gamma family must be strictly larger than zero. we must require α + a > to make this argument. (c) When a, we have E(Y ) β Γ(α + ) Γ(α) β αγ(α) Γ(α) αβ. (d) Note that when a. We must require α + a α + >. (e) Note that when a. ( ) E β Γ (α ) Y Γ(α) E( Y ) E(Y a ), E( Y ) E(Y β Γ ( α + ) Γ(α) ( ) E E(Y a ), Y ) β Γ (α ) (α )Γ (α ) (α )β. We must require α + a α > α >. Remark: E( Y ) is called the first inverse moment of a random variable Y.. Note that ( ) E E(Y a ), Y when a. ( ) E β Γ ( α ). Y Γ(α) We must require α + a α > α >. Note that ( ) E Y E(Y a ), PAGE 8

9 STAT 5 when a. ( ) E Y β Γ (α ) Γ(α) β Γ (α ) (α )(α )Γ (α ) (α )(α )β. We must require α + a α > α >. Remark: E( Y ) is called the second inverse moment of a random variable Y (a) We recognize m Y (t) ( ) 4t as the mgf of a gamma distribution with α and β 4 (provided that t < 4 ). if Y has this mgf, then Y gamma(α, β 4). (b) We recognize m Y (t) 3.t as the mgf of an exponential distribution with β 3. (provided that t < /3.). if Y has this mgf, then Y exponential(β 3.). (c) We recognize m Y (t) exp ( 5t + 6t ) exp (µt + t ) as the mgf of a normal distribution with mean µ 5 and 6. if Y has this mgf, then Y N (µ 5, ). PAGE 9