Question

3. In this question, you will identify the distribution of the sum of independent random variables. I expect you will find that the mgf approach is your friend. (a) Let X and Y be independent Poisson random variables with means A1 and 12, respectively, and let S = X+Y. What is the distribution of S? (b) Let X and Y be independent normal random variables with means Husky and variances 07. 07. respectively, and let S = X+Y. What is the distribution of S?! 4. Consider n independent tosses of a coin with probability of a head equal to p. Let X be the numbers of heads and Y be the numbers of tails. (a) Show that E(X) + E(Y) = n. (b) Show that Cou(X,Y) = -Var(X). (c) Compute Corr(X,Y).

Answer 1

(3) ca) Giner & and Y are independent Poisson distribution with parameters & and iz respectively : Then the paf of x: P(x=x)= x=0,1,2,...,Pdf of Y, P(yzy)= enaz The mgf of! X and Y : . Y! a.x, Mact) - eble-1 of 7, My(t) = è izle-1) Ginen X and Y are independent, ... the mgf of S=X+Yz e McCt)= Mety(t)= M_CE) My A) e é e Chtoelet-02

...... Si er vita e le v od let the litt which is the mgf of a Prisson with parameter 1 tl2. ... By the unique property of mif, sex+y follows Poisson distribution with parameter dit de suppose 11th22 - The probability mass fonction (proof of s, - P(sas)= ets i S= 0, 1, 2,. . . (6 Now X and Y are independent NCMxis 2 and - NCM₂, o 2% respectively. Let S = x+y. we have the probability density function (pdf) of -8: f(x2 (103, HER, HER G 20 The pdf off, feg) = (hy) Samoz oy, yer, Myth, y so The mgf of x, Milt Mt + try2 The mgf jy, My lt) e My Ctit toge Ginen X and Y are independent, Therefore the 2 Ss xty, Melt)=, Mycery CÉ)= My Ct) My CH) - Mutter Myt + teens e tCMx tfly) + mootoyly which is the maf of Normal dis toitution with parameters Mxt My and oto 2. .. by the uniqueness property of not mgf, S=x+y follows Ngomal with theans that hly aud_variance tota The pdf ofs, f(s) = 1 tirotor) e Rotor ress=rostor) e 2oxto?" SER, EXER, AGER potapo

* 4) The given situation can be interpreted as a Binomial distribution. Sop, thens, &~ Binomial (mp) en xy no. of succeeses (no of heads) and these definitely Ya no of failures (no. of tails ) E we know that most if x denotes no of heads, then Y=n-x will be the no of tails : XN Binomilal (mp), meam, Ecx)= np Variance, V(x)= npq aq=14 Y= n-y~ Binomial an, q) means Ely)= ng Variame, VCY)= ng p. 1 ._ca). E{X2key2 mp ung = nCp+9)= in (90€ | - cb con CX, 42= con(x, n-x) = cov(n,x) - Cov(x,x) - 2 - VCX covex,x)=162 i cov(const, X2 = 0 (C) corr(x,y)= CNCX,92 Vex) frex fuerz VAND SUCH) = - - vex2 Jucy)