Question

2. Random variables X and Y have joint probability density function f(x, y) = kry, 0<<1,0 <y <1. Assume that n independent pairs of observations (C,y:) have been made from this density function. (a) Find the k which makes f(x,y) a valid density function, (b) Find the maximum likelihood estimators of a and B. (c) Find approximate variances for â and B.

Answer 1

The necessary condition far a density function to be a valid density function is that , the integration (for continuous r.v.) of the density function for the given range of the variables should be equal to 1.

under the certain regularity condition, ML estimate of a parameter is normally distributed for large n

Toint probability function of Xandy is a foxy) = kx P ouxstosyss - (a) valu of k o for feny to be a valid probability fenation I J fon, y) dx dy = - 5' 5'* xy law dy k J x SJ y dy} dx = 1 Butt B+1 >> {145.09=1 k= (+1) (B+1)

nh (6) MLE of a and B beating the value of k in 0, we get fory) = (2+0 (BH) X Y ;oLX21, OLYLI now, likelihood femetion is one L= 640"-40". A saking log on both sides tog L= nlog (+1) +nlog f41) telogcas y By differentiating wert. d we get la n a logxi - ③ differentiating & wirt ß, we get obot + log Le - @ a for MLE equating to zero lectus a long li' = 0 → The log i - Cat1 = { logxi

> - +1) = Elogue > 12 = -( It togri ) , similencity; equating @ to reso, we get to = -(1+ Ilogy. n (6) Approximate versionees of 2 and ß we know that under cextain regularity condition, mL estimate ê of a parameter o. is assymptatically normally distributed and the assumptatic Capprox.) variance is given by vo) = hEolo log fcxios 2

or vos now. marginal distribution of a gles = (2+1)(B+1) 27 5 yody gexs = 641) (6+1). XP Com). still g(x) = \d+1) 82 - and marginal of Y is hey) = (2-11) (BH) y B Judae thly) = (B+1) yB - 6 now, likelihood for grael Lex, a) = (etun xx taking log, log Loxid) = nlogld+1)+o differentiating wis ta, 2 log LCXid) = 1 11 + log ke again, differentiating world 2 log L(XQ)- - J

similarly, for hiy), we get 22 LLY, B) = - 4 12 12 Now, we take the expectation of ☺46 Jaking expectation of @ 61222 logLexoj] = 2 - 0 taking expectation of a 6 ( 32 log ley,ß)] = - 2 - © butting in D, we get VW -- nuz? Tv (2) = 64413² putting in , we get UCÊ) = { 1 (+132) } (v (B) = (841/2