Question

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(1 point) A normal distribution with mean 0 and standard deviation Vo is sampled three times, yielding values x, y, z. Find the log-likelihood function In L(O) (type theta for 6): In L(0) Find the derivative of the log-likelihood with respect to 0 (type theta for 6): [In LO] Find the maximum likelihood estimator for 0 (note that there is only one positive value): ê=

Answer 1

A with mean o Standard deviation So is If X . Normal dis tribution with mean M and Variance le standard deviation o then p.dif of x is given by 1 e doz (X- 14" - 0022 200 Here we are given, १ co le the e pol(2-0)' +(3-0)' +(2-0)'] to [(xx-03 + (9-03 + (2-03] normal distribution and sampled three times, yield Values x,y,z. f(x) = M= 0, o 1 (x-03 f (OC) To sam (a) Find log - likelihood function Leos likelihood function for of 110) = f(oc).fly) f(z) 1 eto (2-0)* ' et le Co-os? So sam - Joan 0 % (27) 03 (21) Taking loge on In L60)= - 3 Ino - 3 lnan - 112 [(00-03² + 19-03² + (2-03] The o (z-oil 1 Jo San 1 3/2 -312 L(0) - e both sidy

2 Find the derivative of the log. likelihood with respect to o a [ln Leo)] - Mlg де oop - 440.80(x-of++ (7.0° +4-03)(x-08+4 0342-0) +17-03] 0.42(3-0) (-1) +244-0) (-1) + 2(2-0) [++)–468-08+18-03 3 -3-1 -Z2-02 +220 200+209–204+203_202+20?– Je?-0°+220 - y?_0+ 240 -3 20 20² +.20² + 20% _ x²-0²-g²-0² - 2 - 0 2 2 12 3 -3 11 - 4 [60°-38° - (**+7"] ***2"}] -- [ 30- (82)] [ 30- (op+y't 2 o ao In L10)] = - 3 20

© Find the maximum (3) likelihood estimator for o Chote that there is one positive Value) For maximum likelihood estimator for 0 we Solve a [ln Loi] = 0 Do - ( 3 o *21poloplasts] 302 - Cop+y*+7)] 3 2. 20 - [30² - (x² + y² + 2²) 3 o -> - 30² + (x² + y² + 2) = 30 formula ar² tbxtc=0 304 + 30 - (ocº+y+z) - 0 The above equation is quadratic in o. x = -b1 b²-490 2a -3 † 59+4x3 (3e?+y?+28) 2x3 -3 + 9 + 12 (x² + y² +22) 2x3 - 3+ 9+ 36 (oc®+y+z) 2x3

-3 € 3 11+ 4 + 4(x+y+z) 2x3 3 x² + y² + 2² -11 I 9 1 1 1+4. (8°+y*+zº) 2 0 = -1 + 2. 1 + 4 (+9 +2?) 1+ 4 (38*43% +2) or -- But we that the Variance know than is always greater Zero. Here Variance 070 Therefore consider we hegelect the negative value of 0 and the positive value of o for me of o. The mascimum likelihood estimator for o is by. Therefore, given + 1+ 4 +4 (xy+z Ang Q 3