(10) Let X, Y be independent and let X ~ N(a1,b²)and let Y ~ N(a2, b5). Using the properties of mgf, if possible, prove that Z = X + Y is again a normal random variable.The following proofs are provided.(a) The statement of the exercise is, in fact, always false.(c) The statement of the exercise is true when a1 = a2 and b1 = b2 .(c) From the table of mgfs we see that the mgfs of X and Y are¥x(t) = exp{(t – aj)²/2},¥y(t)- exp{(t – a2)²/2}.By the convolution property,Yz(1) = ¥x(t)¥y(1) = exp{(t – (a1 + az))²/2}.%3DThis is an mgf of a normal random variable.(d) From the table of mgfs we see that the mgfs ofX andYare¥x(t) = exp{r?/2},¥y(t) = exp{r²/2}.By the convolution property,Yz(t) = ¥x(t)¥y(t) = exp{2t²/2}.%DThis is an mgf of a normal random variable.(e) From the table of mgfs we see that the mgfs of XandY areYx(t) = exp{a¡t+ (b;t2/2)},Yy(t) = exp{azt + (b;t²/2)}.By the convolution property,Yz(1) = ¥x(t)¥y(1) = exp{(a1 + a2)t + ((b} + b})t² /2)}.This is an mgf of a N(a1 + a2, (b; + b5)),random variable.(a)(b)(c)(d)(e)N/A(Select One)

Given that, X~N(a1,b1^2) and Y~N(a2,b2^2) MGF of X & Y is, ψX(t)=ea1t+b12t22ψY(t)=ea2t+b22t22

(10) Let X, Y be independent and let X ~ N(a1, b;) and let Y ~ N(a2, b;). Using the properties of mgf, if possible, prove that Z = X + Y is again a normal random variable. The following proofs are provided. (a) The statement of the exercise is, in fact, always false. (e) The statement of the exercise is true when aj = a2 and bi = b2. (e) From the table of mgfs we see that the mgfs of X and Y are Yx(t) = exp{(t –- a¡)²/2}, Yy(1) = exp{(t – a2)²/2}. By the convolution property, ¥z(t) = ¥x(t)Yy(1) = exp{(t – (a1 + az))²/2}. This is an mgf of a normal random variable. (d) From the table of mgfs we see that the mgfs of X and Y are Yx(t) = exp{r²/2}, ¥y(1) = exp{r²/2}. By the convolution property, ¥z(t) = ¥x(t)Yy(1) = exp{2r²/2}. This is an mgf of a normal random variable. te) From the table of mgfs we see that the mgfs of X and Y are ¥x(t) = exp{a¡t+ (b¡r²12)}, ¥y(1) = exp{azt + (b;r²/2)}. By the convolution property, ¥z(t) = ¥x(t)\Yy(t) = exp{(a, + a2)t + ((b² + - b²)r²12)}. This is an mgf of a N(a¡ + a2, (b² + b;)) random variable. (a) (b) (c) (d) (e) N/A (Select One)

(10) Let X, Y be independent and let X ~ N(a1, b;) and let Y ~ N(a2, b;). Using the properties of mgf, if possible, prove that Z = X + Y is again a normal random variable. The following proofs are provided. (a) The statement of the exercise is, in fact, always false. (e) The statement of the exercise is true when aj = a2 and bi = b2. (e) From the table of mgfs we see that the mgfs of X and Y are Yx(t) = exp{(t –- a¡)²/2}, Yy(1) = exp{(t – a2)²/2}. By the convolution property, ¥z(t) = ¥x(t)Yy(1) = exp{(t – (a1 + az))²/2}. This is an mgf of a normal random variable. (d) From the table of mgfs we see that the mgfs of X and Y are Yx(t) = exp{r²/2}, ¥y(1) = exp{r²/2}. By the convolution property, ¥z(t) = ¥x(t)Yy(1) = exp{2r²/2}. This is an mgf of a normal random variable. te) From the table of mgfs we see that the mgfs of X and Y are ¥x(t) = exp{a¡t+ (b¡r²12)}, ¥y(1) = exp{azt + (b;r²/2)}. By the convolution property, ¥z(t) = ¥x(t)\Yy(t) = exp{(a, + a2)t + ((b² + - b²)r²12)}. This is an mgf of a N(a¡ + a2, (b² + b;)) random variable. (a) (b) (c) (d) (e) N/A (Select One)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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