STAT Homework 1 - Solutions

Transcription

1 STAT Homework 1 - Solutios Fall 018 September 11, 018 This cotais solutios for Homework 1. Please ote that we have icluded several additioal commets ad approaches to the problems to give you better isight. Problem 1. Suppose we toss a fair coi util we get exactly three heads. Describe the sample space Ω. Let X deote the umber of tosses. Fid probability mass fuctio of X. Solutio 1. The Sample space here is Ω {X 3 : HHH X 4 : THHH, HTHH, HHTH X 5 : TTHHH, THTHH,......} We the have the probability mass fuctio as: ( ) ( ) k 1 1 P(X k) }{{} I first (k 1) tosses there are two ad oly two Hs ( ) ( k 1 1 ) k ( ) 1 k 3 }{{} rem. (k 3) Ts i first (k 1) tosses Problem. Cosider evets A 1, A..., A. Prove that ( ) P A i P(A i ). i1 i1 ( ) 1 }{{} third H o kth toss Solutio. We will proceed by mathematical iductio. ( 1i1 ) Proof. Base case: If 1, the P A i P(A 1 ) 1 i1 P(A i). So, the theorem holds whe 1. Also for the result holds usig Lemma 0.. Iductive hypothesis: Suppose the theorem holds for ( all values ki1 ) of up to some k, k 1. That is we assume that P A i k i1 P(A i).

2 stat homework 1 - solutios Iductive step: Let k + 1. The we have: P ( k+1 i1 ) ( ) k A i P A i A k+1 i1 ( ) ( ) k k P A i + P(A k+1 ) P A i A k+1 i1 ( ) k P A i + P(A k+1 ) i1 k P(A i ) + P(A k+1 ) i1 k+1 P(A i ) i1 i1 }{{} 0, sice it is a probability (by iductive hypothesis) So, the theorem holds for k + 1. By the priciple of mathematical iductio, the theorem holds for all N. (Usig Lemma 0.) Problem 3. Suppose that A ad B are idepedet evets. Show that A ad B c are idepedet evets. 1 Solutio 3. Proof. P(A) P(B c ) P(A) (1 P(B)) P(A) P(A) P(B) P(A) P(A B) P(A B c ) 1 Idepedece is all about what iformatio is cotaied i evets. It is a very strog property to assume but usually doe iitially for mathematical simplicity. As a extesio to this questio are A c ad B c also idepedet? Prove it or provide a couterexample. So evets A ad B c are idepedet i.e. A B c. Problem 4. Show that if P(A) 0 or P(A) 1 the A is idepedet of every other evet. Show that if A is idepedet of itself the P(A) is either 0 or 1. Solutio 4.(a) First part: Proof. Cosider B arbitray evet, suppose P(A) 0, sice A B A, 0 P(A B) P(A) 0. Hece P(A B) 0 P(A) P(B), which implies idepedece. Suppose P(A) 1, the P(A c ) 1 P(A) 0. By the above proof, we have that A c B. Sice (A c ) c A, by problem 3, we get A B. (b) Secod part:

3 stat homework 1 - solutios 3 Proof. Sice we are give that A A this meas that P(A) P(A A) P(A) P(A) (P(A)) P(A) (1 P(A)) 0 From this it follows that P(A) 0 or P(A) 1. Problem 5. Let X have CDF F. Fid the CDF of Y mi{0, X}. Solutio 5. I this case we have F Y (t) : P(Y t) P(mi{0, X} t) 1 P(mi{0, X} > t) 1 P(0 > t) P(X > t) 1 P(t < 0) (1 P(X t)) 1 P(t < 0) (1 F X (t)) F X (t) t < 0 1 t 0 Key poit: max(x, Y) a implies that X a ad Y a. Similarly, mi(x, Y) a implies that X a ad Y a. Problem 6. A radom variable X is stochastically greater tha a radome variable Y if F X (t) F Y (t) for all t ad F X (t) < F Y (t) for some t. Prove that, i this case, ad P(X > t) P(Y > t) for every t, P(X > t) > P(Y > t) for some t. Solutio 6. Proof. We ote that by defiitio the complemetary evets {X t} c {X > t}. As such we have that P(X > t) 1 P(X t) : F X (t). Usig this simple fact we ow proceed to prove the first part as follows 3 : F X (t) F Y (t) t 1 F X (t) 1 F Y (t) P(X > t) P(Y > t) 3 Key poit: We just recogize that the CDF captures left-tail probabilities of a radom variable (i 1D). The complemetary evet is the right tailed probability of the same radom variable. Thereafter we just ote that the probability of a evet ad it s complemet sum to 1 to get a relatioship betwee CDF ad the right-tail probabilities usig Lemma 0.1

4 stat homework 1 - solutios 4 The secod part follows similarly (just keepig carefully otig the strict iequality) F X (t) < F Y (t) t 1 F X (t) > 1 F Y (t) P(X > t) > P(Y > t) Ad thus the required statemets are ow proved. Problem 7. Defie 0 t < F X (t) t t 4 1 t > 4. Prove that F is a valid CDF. Fid the probability desity fuctio. Solutio 7. We will show that F satisfies the required coditios to esure it is a valid CDF (a) lim t F X (t) 0 Proof. We have lim t F X (t) lim t 0 0 (b) lim t F X (t) 1 Proof. We have lim t F X (t) lim t 1 1 (c) F X (t) is odecreasig for all t. Proof. I the cases t < ad t > 4, F X (t) is defied to be costat valued at 0 ad 1 respectively so is o-decreasig over this domai by defiitio. I the iterval [, 4] we ote that for t 1, t [, 4] such that t t 1 we have: F X (t ) F X (t 1 ) t So F X (t) is odecreasig for all t. t 1 t t 1 0 sice t t 1 by assumptio (d) F X (t) is right-cotiuous all t.

5 stat homework 1 - solutios 5 Proof. I the cases t <, t > 4 ad t (, 4) we ote that F X (t) is cotiuous (ad thus right-cotiuous). We just eed to verify right-cotiuity at the ed poits of the iterval (, 4) amely t {, 4}. This is doe as follows: Case: t lim F t X(t) lim t + t 0 F X () Case: t 4 lim F X(t) lim 1 t 4 + t 4 1 F X (4) So F X (t) is right-cotiuous for all t. (e) Fid the PDF. 0 t < f X (t) 1 t 4 0 t > 4. Problem 8. The uiform distributio o [ 3, 3] has desity: f X (x) 1 6 for x [ 3, 3]. Suppose X has this desity. (a) Fid P(X 1). (b) Fid P(0.5 X 1.5). (c) Fid the CDF of Y X. Solutio 8. Note that this is the cotiuous Uiform Distributio o [ 3, 3] ot the discrete versio. (a) Fid P(X 1). This is 0. Proof. For a cotiuous desity the mass is 0 at ay give poit. More formally we have dx 6 1 [x]1 1 0

6 stat homework 1 - solutios 6 (b) Fid P(0.5 X 1.5). This is 1 6. Proof. P(0.5 X 1.5) dx 1 6 [x] (c) Fid the CDF of Y X. Proof. For y [0, 9], F Y (y) P(Y y) ( ) P X y P( y X y ) y y 1 6 [x] y y dx y Hece, the CDF of Y is 0, if y < 0 F Y (y) y 3, if 0 y 9 1, if y > 9 Problem 9. Let (X, Y) have a uiform distributio o the uit circle i the plae. (a) Show that X ad Y are ot idepedet (b) Fid P ( X + Y < 1/4 ). Solutio 9. We proceed as follows: (a) Claim: X, Y are ot idepedet

7 stat homework 1 - solutios 7 Proof. Cosider the evets X > 0.8 ad Y > 0.8. We ote that P((X > 0.8) (Y > 0.8)) 0 sice the two regios are disjoit o the uit circle. However P(X > 0.8) P(Y > 0.8) > 0. Hece, X ad Y ad ot idepedet. 4 (b) Fid P ( X + Y < 1/4 ). Claim: this is 1 4 Proof. This is the probability that X, Y fall i a half circle give that they are joitly uiform o the uit circle. The required probability is simply the ratio of the areas of the circles i this case (sice the joit distributio is cotiuous the boudary of the ier circle has probability measure 0 ad we do t have to be cocered about it s exclusio): 4 Key poit: To disprove idepedece all you sometimes eed is a sigle couterexample to show it does ot hold. Alteratively you ca derive the margials of X ad Y ad show that their product does ot equal the joit desity of (X, Y) ( ) P X + Y < 1/4 π( 1 ) π(1) 1 4 Figure 1: Figure for Q9(a) 5 5 Key poit: Try ad draw a picture ad exploit the geometry of the problem to Problem 10. Let X 1, X,..., X N (µ, σ ). Let T i1 X i. Fid E(T) ad Var(T). Solutio 10.(a) We claim that E[T] (µ + σ ) Proof. Sice the X i s are IID for all it follows that the Xi s are IID for all. We the have E[Xi ] E[X 1 ] fid the quickest solutio. I this case because of the uiform distributio (ad thus uiform volume) i 3D, we are simply cocered with relative D areas to get our required probability E[X 1 ] Var X 1 + (EX 1 ) µ + σ So the we have 6 : E[T] E i1 i1 i1 [ Xi i1 E[X i ] E[X 1 ] ] (µ + σ ) (µ + σ ) (by liearity of expectatio) (sice X i s are idetically distributed) (Sice E[X 1 ] Var X 1 + (EX 1 ) ) 6 Key poit: We did ot rely o idepedece of X i s here to derive the expectatio of T. Simply usig liearity of Expectatio ad Idetically distributed X i s was eough. Always try ad prove statemets with miimal required assumptios

8 stat homework 1 - solutios 8 (b) We claim that Var [T] (µ 4 + 6µ σ + 3σ 4 ) (µ + σ ) Proof. We firstly ote that 7 : ] Var[T] : Var [ Xi i1 7 Key poit: Here we do rely o both idepedece of X i s ad their idetical distributio calculatio to simplify the variace of T i1 Var[X i ] Var[X1 ] i1 (Var[X 1 ]) (sice X i s are idepedet) (sice X i s are idetically distributed) Now Var[X1 ] E[X4 1 ] (E[X 1 ]). From the previous part we kow that E[X1 ] µ + σ. So we just eed to fid E[X1 4] (the fourth momet of X 1 ) ca be foud directly usig the MGF of X 1 as follows: ( ) M X1 (t) e ut+ 1 (σ t ) the MGF for the N (µ, σ ) M (4) ( X 1 (t) 6σ e σ t ) +µt µ + σ σ t + t ( ) 4 e +µt µ + σ t + 3σ 4 e σ t E[X 4 1 ] M(4) X 1 (0) µ 4 + 6µ σ + 3σ 4 +µt (Usig Wolfram Mathematica!) It follows that Var [T] (µ 4 + 6µ σ + 3σ 4 ) (µ + σ ) Appedix: Supportig Lemmas Here we prove some lemmas from the lecture otes that we use repeatedly i the solutios Lemma 0.1. For a evet B we have P(B c ) 1 P(B) Proof. We ote that Ω B B c. Sice B ad B c are disjoit by defiitio we have that 1 : P(Ω) P(B B c ) P(B) + P(B c ) From which we obtai the required result. Lemma 0.. For evets A, B we have P(A B) P(A) + P(B) P(A B)

9 stat homework 1 - solutios 9 Proof. We ote firstly: A (A B c ) (A B) (Writig A as a disjoit uio of sets) P(A) P(A B c ) + P(A B) (takig the probability measure of disjoit sets) P(A B c ) P(A) P(A B) (rearragig terms) By symmetry we also have that P(B A c ) P(B) P(A B). Now fially we ote that we have: A B (A B c ) (A B) (B A c ) (Writig A B as a disjoit uio of 3 sets) P(A B) P(A B c ) + P(A B) + P(B A c ) (takig the probability measure of disjoit sets) P(A) P(A B) + P(A B) + P(B) P(A B) (usig above derived results) P(A B) P(A) + P(B) P(A B) Which is the required result. Refereces