SOLUTION FOR HOMEWORK 4, STAT 4351

1 SOLUTION FOR HOMEWORK 4, STAT 4351 Welcome to your fourth HOMEWORK , which begins our study of chapter 3 and also the lastone before Exam 1. Here we are exploring basics of univariaterandom variables (rv): discreterv, continuous rv, pdf (probability density function), pmf(probability mass function), cdf(cumulative distribution function), calculation of let us look at your Problem A discrete random variableX:=X(w) is defined by its probability massfunction (in the text it is also referred to as the probability distribution but I prefer the pmfas less confusing)f(x),x X. HereXis the range of the random variableX(w) defined forw S. Please remember that bothfandXdefine the pmf of a discrete random variable.(a). Heref(x) = (x 2)/5 andx X={1,2,3,4,5}. Becausef(1) takes on negativevalue, it is not a valid pmf (see Theorem ).(b). Heref(x) =x2/30,x X={0,1,2,3,4}. The function is nonnegative, and4 x=0x2/30 = [1 + 4 + 9 + 16]/30 = is a valid pmf. Please note that you can exclude{0}from the range ofX(w).

2 (c). Heref(x) = 1/5 forx X={0,1,2,3,4,5}. The function is nonnegative but itsums up to 6/5, that is, it is not a valid Problem (a,b). The idea is to find a constantcwhich sums the values of the pmfto 1.(a). Heref(x) =cx,x X={1,2,3,4,5}. Note that the range ofX(w) is given butonly the linear shape of the pmf is given. Because x Xf(x) = 1(1)must hold, we get 1 + 2 + 3 + 4 + 5 = 15 and this yieldsc= 1 (!): the samef(x) =cxbut withx X={1,2,3}isanotherpmf, because here therange is different and, as a result,c= 1/6.(b). Heref(x) =c(5/x) withx X={0,1,2,3,4,5}. Calculate5 x=0(5/x) = and this implies that something is wrong here. Let us remove 0from the range. Then5 x=1(5/x) = 5[1 + 1/2 + 1/3 + 1/4 + 1/5] =:A,andc= 1/A(I skip calculation ofA).13. Problem Note that the cdf is a complete and unique description of a randomvariable. By analyzing the given cdfF(x) we find that the rv at hand is discrete with therangeX={1,4,6,10}and the corresponding pmf isf(1) :=P(X= 1) = 1/3,f(4) :=P(X= 4) = 1/2 1/3 = 1/6,f(6) =P(X= 6) = 5/6 1/2 = 1/3 andf(10) =P(X=10) = 1 5/6 = 1/6.

3 Please check that the total is :(a)P(2< X 6) =P(X= 4) +P(X= 6) = 1/2. Note that the complementary eventis{X= 1} {X= 10}, so the probability of interest is equal to 1 minusP(X= 1) minusP(X= 10).(b)P(X= 4) = 1/6.(c) This was done in the Introduction to my SOLUTION (seef(x)).4. Problem (a) First, I use the law of total probability, and then definition of the cdf:P(X > x) = 1 P(X x) = 1 F(x) =:G(x).Note that these relations hold for anyxand any random variable(!) Also,G(x) is called thesurvivor (distribution) function of the rvX(in biostat/quality-control applications it is theprobability that a patient/item will live/survive after timex).(b) Similarly, for anyxP(X x) = 1 P(X < x)and this holds for anyxand anyX. However, ifx=xiis the atom (the point from therange of a discreteX(w) or the point of jump of the cdf different words for the samenotion), andXis discrete with the range{xi, i= 1,2, .. , n}then for any 2 i nP(X xi) = 1 P(X xi 1) = 1 F(xi 1).In the last equality I used definition of the Problem First of all, look how I am writing the pdf (probability density function)f(x) =cx 1/2I(x (0,4))whereI(x A) is the indicator (or the indicator function) of the eventAmeaning thatthe indicator equal to 1 ifxbelongs toAand it is equal to zero otherwise.

4 Note that mydefinition of the pdf and the book s one are we can consider the problems.(a): To find the value ofcwe need to remember that the pdf is integrated to 1 over( , ), that is, f(x)dx= 1.(2). For our specific density f(x)dx= 40cx 1/2dx=c[2x1/2]40=c[(2)(2) 0] = to (2), we getc= 1/4.(b) By definition of the probability of an eventAand pdf of a continuous random variableX,P(A) =P(X A) = Af(x)dx.(3)Using (3) we get:P(X <1/4) = 1/4 f(x)dx= 1/4 (1/4)x 1/2I(x (0,4))dx= 1/40(1/4)x 1/2dx= (1/4)[2x1/2]1/40= (1/2)[1/2 0] = 1 ,P(X >1) = (1/4) 41x 1/2dx= (1/4)[2x1/2]41= (1/2)[2 1] = 1 the way, what are the probabilities:P(X >5) =?,P(X= 3) =?.6. Problem We can write the pdf asf(z) =kze z2I(z >0).Now let us findk. Using (2) we get1 = f(z)dz= 0kze I see a nice change of the variablex=z2which simplifies the integration. Let us dothis: 0kze z2dz=k 0(1/2)e z2dz2= (k/2) 0e xdx= (k/2)[ e x] 0= (k/2)(1) = yieldsk= about graphic it is 0 forz 0, then it is increasing up to pointz0such that thefirst derivativef (z) :=df(z)/dzis equal to zero, that is,f (z0) = 0, and then it goes downto zero.

5 Let us findz0. We have:f (z) = 2[e z2 z(2ze z2)].Then the derivative is equal to zero at pointz0= 1/2. This is the point of maximum(check this by looking at the second derivative or from the graphic).7. Problem Well, the question is about the cdf (cumulative distribution function).Well,F(x) = x f(z)z.(4)3By an elementary calculation we get thatF(x) = 0I(x 0) + (1/3)xI(0< x 1) + (1/3)I(1< x 2)+[(1/3) + (1/3)(x 2)]I(2< x 4) +I(x >4).Please check that the cdf is not decreasing, 0 atx= and 1 atx= . Further, becausethe random variable is continuous, the cdf is also Problem For the given cdf we see that it is continuousand its support is [ 1,1](the set where the corresponding density is positive). Then:(a)P( 1/2< X <1/2) =F(1/2) F( 1/2) = 3/4 1/4 = 1/2.(b) Similarly,P(2< X <3) =F(3) F(2) = 1 1 = is becauseXtakes on values larger than 1 with zero Problem Similarly to the previous problem (and hereactually due to the defini-tion of the cdf)P(Y 5) =F(5) = 1 9/25 = 16 (Y >8) = 1 P(Y 8) = 1 F(8) = 1 (1 9/64) = 9 Problem Here we should be very accurate with equalities because with mixedrv we may haveP(X=x) being zero or positive depending onx.

6 (a)P(Z= 2) =F( 2) F( 2 ) = 1/4 0 = 1 (x ) denotes the value ofF(z) to the left of pointz=x.(b)P(Z= 2) =F(2) F(2 ) = 1 6/8 = 1 note thatz= 2 andz= 2 are the discrete components of the random variable (thecdf has jumps at these points).(c)P( 2< Z <1) =P(Z <1) P(Z 2) =F(1 ) F( 2) = 5/8 2/8 = 3/8.(d)P(0 Z 2) =P(Z 2) P(Z <0) =F(2) F(0 ) = 1 4/8 = 1 hope that I made no mistakes. Agree?4