MATH2715: Statistical Methods

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1 MATH275: Statistical Methods Exercises III (based on lectres 5-6, work week 4, hand in lectre Mon 23 Oct) ALL qestions cont towards the continos assessment for this modle. Q. If X has a niform distribtion oer the interal (0,), proe that U = log X has an exponential distribtion with mean one. Hint: X has probability density fnction f X (x) = for 0 < x <. Find the probability density fnction of U. Q2. (a) If X has a niform distribtion oer the interal (0, ), obtain the probability density fnction of U = X 2. (b) If X has a niform distribtion oer the interal ( 2,+ 2 ), obtain the probability density fnction of U = X 2. Hint: Is this a mapping? Q3. A non-negatie random ariable X has probability density fnction f X (x) = e x, x > 0. Derie the probability density fnction f U () of U = log X. Sketch this fnction f U (). Hint: Don t forget that yo shold always specify the range of a probability density fnction. To sketch f U () yo cold look at what happens for large and small and check for trning points. Yo cold also try sbstitting different nmerical ales for. Q4. Sppose that X has an exponential distribtion with parameter λ = 2. (a) Find the probability density fnction of U = + X. (b) Obtain the mean of U. Hint: Recall that if Z N(0,) with probability density fnction f Z (z) = e 2 z2, defined for < z <, then E[Z 2 ] = Var[Z] + {E[Z]} 2 = =. (c) Simlate 000 ales of X and hence generate 000 ales of U. What is the mean of yor simlated U ales? Hint: R commands: x=rexp(000,0.5) # Pts 000 exponential(lambda=0.5) ales into x. =sqrt(x) Q5. Sppose that random ariables X and Y are independent standard normal random ariables. (a) If U = X 2 + Y 2 and V = X/Y, obtain the joint probability density fnction f UV (,). Hint: To find the Jacobian notice that J = (x, y) (, ) = (,) (x, y). Is the mapping (x,y) (,) a mapping? (b) Proe that U and V are independent. Hint: Show that f UV (,) factorises as f U ()f V () where f U () and f V () are recognisable probability density fnctions. 24

2 Backgrond Notes: Lectre 5: Fnctions of a random ariable Example Power of wind trbines 24 For a wind with instantaneos speed V the power is P = 2 λv 3 where λ is the density of the air 25. Typically windspeeds are aeraged oer a day and the aerage daily windspeed U is sch that U can be modelled as a normal distribtion. The power P aailable to a wind trbine typically satisfies P U 5/2. Example Income in UK 26 Incomes U are sch that X = log U is well-modelled by a normal distribtion; see figre 0. Freqency per interal of Millions per interal of width one nit Income before tax (ponds) log 0 (Income before tax) Figre 0: (left) histogram of income before tax , (right) histogram of log 0 (income). Reasons for transforming random ariables A common goal in transforming ariables is to indce symmetry and homogeneity in the error distribtion a transformation designed to achiee one prpose...often also helps to achiee another. 28 Backgrond Notes: Lectre 6: Fnctions of two random ariables Jacobian of a transformation Consider the transformation = (x, y) and = (x, y) which maps R xy in the (x,y) plane into the rectanglar region R with area in the (,) plane; see figre. 24 Sorce: Haslett, J. and Raftery, A.E. (989) Space-time modelling with long-memory dependence: assessing Ireland s wind power, Applied Statistics, 38, Imagine a nit area in space and let λ be the density of the air. For a wind with speed blowing perpendiclar to the area, a olme of air eqal to will be moed in nit time. The mass of air that moes will eqal m = λ. The kinetic energy of this moing mass eqals 2 m2 = 2 λ3 and per nit time the power deeloped will eqal 2 λ3. 26 Sorce: Inland reene, 27 Sorce: Kettl, S. (99) Acconting for heteroscedasticity in the transform both sides regression model, Applied Statistics, 40, Sorce: Kendall, M. and Start, A. (976, p.95) The Adanced Theory of Statistics, olme III (3rd edition), Griffin, London. 25

3 Y A (x,y) C R xy B V + R X Figre : Mapping R xy to R. + U The determinant J = (x,y) (,) = is referred to as a Jacobian 29 and the area R xy can be shown 30 to satisfy R xy J. It can be shown that (, ) (x,y) = = ( ) (x,y) = (, ) J. For example, if = x/y and = y, then x = and y = so that (x,y) (,) = (, ) (x,y) = /y x/y 2 = /y = /. 0 0 = while 29 Carl Jacobi (804-85) was a German mathematician who worked extensiely with determinants. 30 In figre, let point A map to (, ) and B map to ( +, ) so A has x-coordinate x(, ) while B has x-coordinate x(+, ) where x = x(,) and y = y(, ) denotes the inerse transformation from the (, ) plane to the (x,y) plane. Since = lim x( +, ) x(,) 0 it follows that the x-difference of A and B is B is y( +, ) y(, ). Hence the ector AB satisfies AB = x( +, ) x(, ). Similarly the y-difference of A and ««i + j where i and j are nit ectors in the x and y directions respectiely. Similarly ««AC = i + j. For small and, R xy will be approximately a parallelogram with area AB AC where denotes a ector prodct. Ths the area of R xy is gien by AB AC (x,y) = (, ). 26

4 Frther reading for lectres 5 and 6 Rice, J.A. (995) Mathematical Statistics and Data Analysis (2nd edition), section 2.3, 3.6, 4.. Hogg, R.V., McKean, J.W. and Craig, A.T. (2005) Introdction to Mathematical Statistics (6th edition), sections.7, 2.2, 2.7, 3.3. Larsen, R.J. and Marx, M.L. (200) An Introdction to Mathematical Statistics and its Applications (5th edition), sections 3.8, 7.3. Miller, I. and Miller, M. (2004) John E. Frend s Mathematical Statistics with Applications, sections 6.3, 7.3,

5 MATH275: Statistical Methods Worked Examples III Worked Example: A random nmber between 0 and is a ale niformly distribted between 0 and. Sppose yo hae aailable a seqence of sch random nmbers. Sggest how yo cold generate a seqence of ales haing an exponential distribtion with parameter λ, mean µ = /λ. Answer: Here X niform(0,) so f X (x) = for 0 < x <. exponential(λ) so that f U () = λe λ for > 0. Ths f U () = f X (x) dx d = λe λ, > 0, We want to hae U so we reqire dx d = λe λ x = e λ. The reqired transformation is ths = log x. λ Worked Example: If X niform(0,), find the probability density fnction of U = X. Answer: The mapping x is. if x, f X (x) = for 0 < x < and F X (x) = x if 0 < x <, 0 if x 0. Range of U: 0 < x < = 0 < <. Using cmlatie distribtion fnction: { } F U () = pr{u } = pr X = pr { X 2} = F X ( 2 ) = 2 for 0 < <. f U () = df U() = 2 for 0 < <. d Direct approach: = x so x = 2 and dx d = 2. Ths f U() = f X (x) dx d = 2. The reslt of a simlation 3 of 2000 ales U = X is plotted in figre 2(left) and the histogram shows the U ales hae a probability density fnction f U () = 2. Worked Example: If X niform(,), find the probability density fnction of U = X 2. 3 Sitable R code is: x=rnif(2000) # Pt 2000 niform(0,) ales into x. =sqrt(x) # =sqrt(x). hist(x,breaks=00) # Histogram of x with 00 interals. hist(,breaks=00) # Histogram of with 00 interals. 28

6 Freqency per interal of width Freqency per interal of width Figre 2: (left) histogram of 2000 simlated U = X ales for X niform(0,) and probability density fnction of U; (right) histogram of 2000 simlated U = X 2 ales for X niform(,) and probability density fnction of U. Answer: f X (x) = 2 for < x < and F X(x) = Range of U: Since < x <, range of U satisfies 0 < <. Using cmlatie distribtion fnction approach: if x, 2 + 2x if < x <, 0 if x. F U () = pr{u } = pr { X 2 } = pr { < X } = F X ( ) F X ( ) =. f U () = df U() d = 2 for 0 <. Direct approach: The mapping x is not. Both x = and x = + gie the same -ale. Ths split the range of X into {x < 0} and {x > 0}. Since = x 2, then x = ± and dx d = ± 2 2. f U () = f X (x = ) dx d +f X (x = + ) dx x= d = x= = 2. The reslt of a simlation 32 of 2000 ales U = X 2 is plotted in figre 2(right) and the histogram shows the U ales hae probability density fnction f U () = /2. Worked Example: Sppose X N(0, ). Use the cmlatie distribtion fnction approach to obtain the distribtion of U = X 2. Answer: Here F U () = pr{u } = pr { X 2 } = pr { X } ; 29

7 f X (x) X ~ N(0,) = x 2 x = - /2 0 x = + /2 x - /2 0 /2 x Figre 3: (left) U if X ; (right) shaded region is pr{ X }. see figre 3. Since X N(0,), so pr{a < X b} = F U () = b a f X (x)dx = b a e 2 x2 dx e 2 x2 dx = 2 e 2 x2 dx, 0 sing symmetry. Differentiating nder the integral 33 gies f U () = df U() d = 2 e 2 ( ) 2 ( ) 2 2 = e 2, > Sitable R code is: x=rnif(2000,-,) # Pt 000 niform(-,) ales into x. =x^2 # =x^2 hist(,breaks=c(0:00)*0.0)) # Histogram of with 00 interals. 33 Sppose that G() = Z b a g(x,) dx where a and b are fnctions of and g(x,) is a fnction of x and. Then dg() d = g(b,) db Z da b g(a,) d d + g(x, ) dx. a This is sometimes known as the Liebnitz integral rle. Z 2 Example: G() = e x dx dg() Z 2 = e 3 2 (e 0) + xe x dx. Z d Example: G() = e x f(x)dx dg() Z = xe x f(x)dx on writing g(x,) = e x f(x) and noting that d the integration limits Z do not depend on. b Example: G() = e 2 x2 dx dg() = e 2 b2 db with g(x,) = e 2 x2 and noting that g d d = 0. Z To proe the formla for differentiation nder the integral sppose G (x,) = g(x,)dx so G (x, ) = g(x,). Then 2 G (x, ) = 2 G (x,) «G (x, ) = «G (x,) = g(x,). Z Integrating with respect to x gies G (x,) g(x,) = dx. This reslt implies: () Z Z Z g(x,) g(x,)dx = dx, and (2), as a definite integral, G (b, ) G (a, ) b g(x, ) = dx. 30 a

8 Worked Example: Use the cmlatie distribtion approach to derie the probability density fnction of U = e X when X N(µ,σ 2 ). Answer: Here X N(µ,σ 2 ) and U = e X, = e x. Ths pr{u } = pr { e X } { = pr{x log } = pr Z log µ } ( ) log µ = Φ σ σ where Z = X µ σ N(0,) and Φ(z) = pr{z z}. Hence F U () = log µ z0 = σ z= e 2 z2 dz and, recalling differentiation nder an integral with dz 0 d = σ, f U () = df U() d = e 2 z2 0 dz 0 d = exp { (log µ) 2 2σ 2 } σ, > 0, Worked Example: If a random ariable X has probability density fnction f X (x) and cmlatie distribtion fnction F X (x), proe that U = F X (x) niform(0,). Answer: Since U = F X (x), then d dx = df X(x) = f X (x) so f U () = f X (x) dx dx d = f X(x) d =. dx Since 0 F X (x) it follows that 0 and so U niform(0,). Worked Example: If X,Y ind niform(0, ), find the probability density fnction of U = X/Y. Answer: Since X and Y are independent, f XY (x,y) = f X (x)f Y (y) = for 0 < x,y <. Let = x/y, = y. Inerse transformation is y =, x = y = and the Jacobian is J = (x,y) (,) = = 0 =. Considering now the definite integral G() = dg() d = G (b, ) b Z b a g(x,) dx, it follows that G() = G (b,) G (a, ). Hence db d + G (b, ) G (a, ) da a d G (a, ) = g(b, ) db d + G (b, ) g(a,) da d G (a,) = g(b,) db d g(a,)da d + ( ) Z b a g(x, ) dx. The reslt ( ) follows from obtaining the total deriatie of G (b, ) and G (a, ). Recall that for a fnction f(x, y) f(x, y) f(x, y) with x = x() and y = y() both fnctions of, then = f(x, y) +. Writing f(x, y) = G (b, ) gies G (b, ) = G (b, ) b b + G (b, ) = G (b, ) b b + G (b, ). Similarly G (a, ) can be obtained. 3

9 Clearly this is a mapping 34. Ths f UV (,) = f XY (x,y) J = =. Range of (U,V ): Clearly = x/y satisfies 0 < <. 0 < y < = 0 < <. 0 < x < = 0 < < = 0 < < /. For to satisfy both constraints 0 < < and 0 < < / we need 0 < < min(,/). Ths if 0 < <, then 0 < < whereas if > then 0 < < /. This gies the region (,) where f UV (,) is defined. Figre 4(left) shows the region where (x, y) is defined while figre 4(right) shows the corresponding region in the (,) plane. The defined region in the (,) plane depends on whether < or >. It can be seen that the mapping (x,y) (,) is a mapping. Figre 5(left) shows 2000 simlated ales of (,). y =0.5 =0.5 = =2 = =0.5 =0.5 = =2 = x Figre 4: showing region where (left) (X,Y ) defined, (right) (U,V ) defined, together with lines showing = 2,,2,4 and = 2. A different way to obtain the range of (,) is to notice that = x/y implies y = x/. This is the eqation of a straight line with slope /; see figre 4(left). (a) If 0 < <, then this line passes throgh the origin (x,y) = (0,0) and has slope (/) >. Ths this line intersects the nit sqare defining the (x,y)-region at (x,y) = (,). Along this line y takes all ales between 0 and. Hence for 0 < <, it follows that 0 < <. (b) Now consider the case >. In this case the line y = x/ has slope (/) <. This line again passes throgh the origin (x, y) = (0, 0) and intersects the nit sqare defining the (x, y)-region at (x,y) = (,/). Along this line y only takes ales between 0 and /. Hence for the case >, it follows that 0 < < /. Probability density fnction of U: Notice that the region of integration with respect to depends on the ale of. d = 2 if 0 <, f U () = f UV (,)d = =0 / d = 2 2 if >. =0 34 Each (x, y) pair maps to a single (, ) pair. By inspection a particlar (, ) pair cn only originate from a specific (x,y) pair. 32

10 A simlation stdy generated 2000 ales of (U,V ), as shown in figre 5(left). 35 The histogram in figre 5(right) show the U ales hae a distribtion with a long tail. Concentrating on the interal 0 < < 5 it is seen that the histogram of ales is approximately constant for < and then decays as / 2 for > Freqency per interal of width Figre 5: (left) 2000 random ales of (,); (right) histogram of simlated ales and probability density fnction of U. Worked Example: If X, Y are niformly distribted inside the nit circle, find the probability density fnction of U = X/Y. Answer: f XY (x,y) = π for x2 + y 2 <. Let = x/y, = y. Inerse transformation is y =, x = y = and the Jacobian is J = (x,y) (,) = = 0 =. The mapping is a mapping eerywhere except for when y = 0. Since these points map to ±, they do not case a problem. Hence f UV (,) = f XY (x,y) J = π. Range of (U,V ): Notice = x/y satisfies < < whilst x 2 + y 2 < = < = 2 < + 2 = + 2 < < Sitable R code is: x=rnif(2000) # Pt 2000 niform(0,) ales into x. y=rnif(2000) # Pt 2000 niform(0,) ales into y. =x/y; =y # =x/y, =y. plot([<5],[<5],xlab="",ylab="") # Plot (,) locations for cases with <5. hist(,breaks=000) # Histogram of with 000 interals. hist(,breaks=c((0:50)*0.,max()+),xlim=c(0,5)) # Specify break locations. 33

11 Probability density fnction of U: f U () = f UV (,)d = = = =0 0 π d = [ 2 π d = 2 = + 2 ] + 2 =0 ( ) π d + = + 2 =0 π( + 2 ). π d For a simlation stdy 36 of 2000 ales of (U,V ), figre 6(left) shows the distribtion of (X,Y ) whilst figre 6(right) shows the distribtion of (U,V ). A histogram of U wold show that the Cachy distribtion U has a long tail. y x Figre 6: (left) 2000 points niformly distribted oer nit circle, (right) mapping to (, )-plane shows density of (U,V ) increases as increases, so f UV (,). Worked Example: If X has a standard normal distribtion and Y, independent of X, has probability density fnction { ye 2 y2 if y > 0, f Y (y) = 0 if y 0, 36 Sitable R code is: xx=rnif(5000,-,) # Pt 5000 niform(-,) ales into xx. yy=rnif(5000,-,) # Pt 5000 niform(-,) ales into yy. x=xx[xx^2+yy^2<] # Set x and y to only contain ales y=yy[xx^2+yy^2<] # of xx,yy lying in region xx^2+yy^2<. x=x[:2000] # Make x contain first 2000 ales of x. y=y[:2000] # x and y now contain jst 2000 points. =x/y # =x/y. =y hist(,breaks=000) # Histogram of with 000 interals. hist(,breaks=000,xlim=c(-5,5)) plot(xx,yy,xlab="x",ylab="y") # Plot shows (xx,yy) points niformly. plot(,,xlim=c(-5,5),xlab="",ylab="") # Plot shows (,) distribtion. 34

12 show that U = X/Y has probability density fnction f U () = What is E[U]? Answer: As X and Y are independent, their joint probability density fnction is 2( + 2, < <. ) 3/2 f XY (x,y) = f X (x)f Y (y) = e 2 x2 ye 2 y2 = ye 2 (x2 +y 2), < x <, y > 0. For the transformation = x/y, = y, so x = and y =, the Jacobian is J = (x,y) (,) = = 0 =. Inspection shows this is a mapping. Hence, with x 2 + y 2 = () = 2 ( + 2 ), f UV (,) = f XY (x,y) J = ye 2 (x2 +y 2) = 2 e 2 (+2 ) 2. Range of U and V : Since < x < + and y > 0 it follows that < < + and > 0. y = =0 =0.5 = = = =0 =0.5 = = x Figre 7: showing region where (left) (X,Y ) defined, (right) (U,V ) defined, together with lines showing =, 2,,2. Recall that < x < while y > 0. Probability density fnction of U: f U () = =0 f UV (,)d = =0 2 e 2 (+2 ) 2 d. Approach : Notice if W N(0,σ 2 ), then E[W] = 0 so σ 2 = E[W 2 ] = Hence f U () = w=0 =0 Approach 2: σ 2 w2 e w2 2σ 2 dw = σ2 f U () = 2 e 2 (+2 ) 2 d = σ 2. Now pt σ2 = so that + 2 =0 w= σ 2 2 e 2 2σ 2 d = σ3 2 = Pt z = 2 ( + 2 ) 2 so dz d = ( + 2 ), = z=0 2z + 2 e z σ 2 w2 e w2 2σ 2 dw. 2( + 2 ) 3/2. 2z ( + 2 ), d = dz 2z( + 2 ) = π( + 2 ) 3/2 z=0 dz 2z( + 2 ) z 2 e z dz 35

13 Recall the gamma fnction Γ(k) = z=0 z k e z dz with Γ ( ) 3 2 = 2 Γ( ) 2 = 2 π = π( + 2 ) 3/2 Γ ( ) 3 2 = 2( + 2, < <. ) 3/2 Mean of U: E[U] = 0 as f U () is symmetric abot zero. Worked Example: If X niform(, ) and Y niform(0, ) are independent random ariables, what is the probability density fnction of U = XY? Answer: f X (x) = 2 for < x < and f Y (y) = for 0 < y <. As X and Y are independent, their joint probability density fnction is f XY (x,y) = f X (x)f Y (y) = 2 for < x < and 0 < y <. The transformation = xy and = y has inerse x = / and y =. The Jacobian J satisfies J = (x,y) (,) = (x,y) (,) = = / / 2 0 It is perhaps easier to ealate the Jacobian J by considering instead (, ) (x,y) = = y x 0 = y. =. The Jacobian then satisfies J = (x,y) (,) = ( ) (,) = (x, y) y. Inspection shows this is a mapping. Hence f UV (,) = f XY (x,y) J = 0.5 y = 2. Range of U and V : 0 < y < = 0 < <. < x < = y < xy < +y = < < < + < +. If > 0, then 0 < < <. If < 0, then < < < 0 so < <. Figre 8(left) shows 000 simlated ales of (,). Probability density fnction of U: Notice the region of integration with respect to depends on the particlar ale of. f U () = f UV (,)d = = 2 d = [ 2 log ] = = 2 log( ) if < < 0, 2 d = [ 2 log ] = = 2 log() if 0 < <. = Clearly f U () = 2 log( ) = log( / ) for < < A more direct approach wold notice that < < if > 0 whilst < < if < 0. Ths we hae < < so that f U() = Z = f UV (, ) d = 2 log( ). 36

14 Freq. per interal Figre 8: (left) 000 simlated ales of (, ); (right) histogram of ales and probability density fnction of U. A simlation stdy 38 yields a plot of the (U,V ) region in figre 8(left). The histogram in figre 8(right) show the U ales together with the probability density fnction of U. Worked Example: Sppose that X and Y are independent standard normal random ariables. Let U = X 2 + Y 2 and V = Y. Obtain the marginal probability density fnction of U. Answer: As X and Y are independent, their joint probability density fnction is f XY (x,y) = f X (x)f Y (y) = e 2 x2 e 2 y2 = e 2 (x2 +y 2 ) for < x < + and < y < +. The transformation = x 2 + y 2 and = y satisfies x = ± 2 and y =. It is ths not a transformation; both (x,y) and ( x,y) map to the same ale (,). It is easier to ealate the Jacobian J by considering instead (, ) (x,y) = = 2x 2y 0 = 2x. The Jacobian then satisfies J = (x,y) (,) = ( ) (,) = (x, y) 2x = ± Sitable R code is: x=rnif(000,-,) # Pt 000 niform(-,) ales into x. y=rnif(000) # Pt 000 niform(0,) ales into y. =x*y # =x*y. =y # =y. plot(,,xlab="",ylab="",main="") # Plot of simlated (,) ales. hist(,breaks=c(-50:50)*0.02,xlab="",ylab="freq. per interal 0.02",main="") cre(000*0.02*(log(sqrt(/abs(x)))),-,,add=true) # Add pdf of U. 37

15 Since both (x,y) and ( x,y) map to the same ale (,), f UV (,) = f XY (x,y) f XY ( x,y) 2 2 = e 2 (x2 +y 2) e 2 (( x)2 +y 2) 2 2 = 2 e 2 (x2 +y 2) 2 2 = e 2 2. Range of U and V : It crdely appears that > 0 and < < +. More careflly, 2 = x 2 < so < < +. Alternatiely, consider the geometrical interpretation of the transformation. Ths = x 2 + y 2 defines a circle in the x-y plane with centre (0,0) and radis. For this circle, the y ales range between + and. Ths with = y we hae < < +. Figre 9 shows the reslts of plotting simlated ales of (X,Y ) in the - plane Figre 9: showing (U,V ) plotted for 000 random pairs (X,Y ) where U = X 2 + Y 2, V = Y and X, Y are independent standard normal random ariables. 39 R code is: x=rnorm(000) # Generate 000 ales x as N(0,). y=rnorm(000) =x 2+y 2; =y # The mapping sed. plot(,) # Plot against. 38

16 Probability density fnction of U: Here 40 f U () = f UV (,)d = + = + = e 2 = e 2 for > 0, so U exponential(λ = 2 ). e 2 2 d = d 2 [ sin ] + = = e 2 [ 2 π ( 2 π)] = 2 e 2 Worked Example: Sppose that X and Y are independent random ariables each haing a niform distribtion on the interal (0,). Let U = X 2 + Y 2 and V = X/Y. Obtain the marginal probability density fnction of V. Show that the marginal probability density fnction of U is a constant for. Answer: f X (x) = for 0 < x < and f Y (y) = for 0 < y <. As X and Y are independent, their joint probability density fnction is f XY (x,y) = f X (x)f Y (y) = for 0 < x < and 0 < y <. The transformation = x 2 + y 2 and = x/y satisfies x = y so = 2 y 2 + y 2. It is easier to ealate the Jacobian J by considering instead (, ) (x,y) = = 2x 2y /y x/y 2 = 2(x2 /y 2 ) 2 = 2(x 2 + y 2 )/y 2. The Jacobian then satisfies J = (x,y) (,) = ( ) (,) = (x, y) Inspection shows this is a mapping. Hence f UV (,) = f XY (x,y) J = Range of U and V : Clearly 0 < < 2 and 0 < <. y 2 2(x 2 + y 2 ) = 2( + 2 ). 2( + 2 ). More careflly, consider the region defined by = x 2 + y 2 in the x-y plane. This is a qarter circle only if 0 < < and along this qarter circle = x/y can take any ale in the interal 0 < <. Howeer, for < < 2 the qarter circle intersects the nit sqare making p the allowed (x,y)-region at (x,y) = (,) and at (x,y) = (, ) since = x 2 + y 2. For this ale of, the range of satisfies < < (/ ). Figre 20(left) shows the case = 0.6 for which 0 < < and also the case =.2 for which < < The lower limit corresponds to the case (x, y) = (0.447, ) and the pper limit to the case (x, y) = (, 0.447). The ineqality < gies < + 2 and this can only hold if < since < 2, so 0 < < + 2 if 0 < <. Similarly the ineqality < (/ ) gies < + 2 and this can only hold if > since < 2, so 0 < < + 2 if < <. Z 40 Recall that dx a2 x = x 2 sin a. 39

17 y =0.6 = Freq. per interal x Figre 20: (left) showing range of for different in (x,y)-plane; (centre) region where (,) is defined in (,)-plane; (right) histogram of ales and probability density fnction of V. An alternatie way to obtain the range of for gien is as follows. Since = x/y it follows that y = x/ is the eqation of a straight line throgh the origin (x,y) = (0,0) with slope /. (a) If 0 < <, then the line has slope (/) >. This line intersects the nit sqare making p (x,y)-region at (x,y) = (,) so that = + 2. In this case takes ales in the interal 0 < < + 2. (b) If >, then the line has slope (/) <. This line intersects the nit sqare making p the (x,y)-region at (x,y) = (,/) so that = + 2. In this case takes ales in the interal 0 < < + 2. Probability density fnction of V : Depending on the particlar ale of, the integration limits for change. + 2 [ ] f V () = f UV (,)d = =0 2( + 2 ) d = + 2 2( + 2 = if 0 < <, ) = [ ] 2( + 2 ) d = + 2 2( + 2 = ) 2 2 if < <. =0 A simlation stdy 4 yields the histogram in figre 20(right) which shows the ales together with the probability density fnction of V. Figre 20(centre) shows the region where (, ) is defined. Probability density fnction of U: In the case <, f U () = f UV (,)d = =0 =0 2( + 2 ) d = [ 2 tan () ] 0 = π 4. Strictly to answer the qestion posed it is only necessary to show that the integration does not inole. It was not necessary to ealate the integral! 4 Sitable R code is: x=rnif(2000) # Pt 000 niform(0,) ales into x. y=rnif(2000) # Pt 000 niform(0,) ales into y. =x^2+y^2; =x/y plot([<4],[<4],xlab="",ylab="") # Plot (,) ales for <4. # Add histogram of ales for <0. hist([<0],breaks=c(0:00)*0.,xlab="",ylab="freq. per interal 0.",main="") lines(c(0,),c(00,00)) # Add pdf of V for <. cre(2000*0.*(/(2*x^2)),,0,add=true) # Add pdf of V for >. 40

18 MATH275: Statistical Methods Examples Class III, week 4 The qestions below will be looked at in Examples Class III held in week 4. Q. If X is niformly distribted oer ( 2 π, ), show that U = tan X has a Cachy42 distribtion. 43 Q2. Random ariables X and Y are independent and each has an exponential distribtion with parameter λ. Consider the transformation U = X/Y and V = Y. Obtain the probability density fnction of U = X/Y. 42 Agstin-Lois Cachy ( ) was a French mathematician who became associated with what is now known as the Cachy distribtion throgh his work on the density in 853. Properties of the Cachy distribtion had howeer been discssed earlier by Poisson in 824 whilst examining properties of the Least Sqares method. 43 Link with lectres: in the lectres the ratio of two independent N(0,) ariables gae a Cachy distribtion. Ths if X, Y ind N(0,) and (X, Y ) defines a point in the plane, then in polar co-ordinates this point satisfies (R,Θ) = ( X 2 + Y 2,tan (X/Y )) where Θ is measred clockwise from the y-axis; compare this with qestion 5 of exercises III. Hence tan Θ = X/Y Cachy. In the lectres the random ariable Θ wold hae a niform( π, π) distribtion and the mapping (X, Y ) (U = X 2 + Y 2, V = X/Y ) is not a mapping. In this examples class qestion, howeer, the range of Θ is ( π, π) and this ensres we hae a mapping