(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. (c) The statement of the exercise is true when a1 = az and bi = bz. (c) From the table of mgfs we see that the mgfs of X and Y are Vx(t) = exp{(t – a,)* /2}, Vy (t) = exp{(t – az)² /2}. By the convolution property, Vz(t) = ¥x(t)¥y(t) = 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 Vx(t) = exp{t* /2}, Vy (t) = exp{t* /2}. By the convolution property, Vz(t) = ¥x(t)¥y (t) = exp{2t² /2}. This is an mgf of a normal random variable. (e) From the table of mgfs we see that the mgfs of X and Y are Vx(t) = exp{a,t + (b;t² /2)}, Vy (t) = exp{azt + (b;t² /2)}. By the convolution property, ¥z(t) = ¥x(t)¥y (t) = exp{(a1 + az)t + ((b} + b3)t² /2)}. This is an mgf of a N(a, + a2, (b + b;)) random variable.
(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. (c) The statement of the exercise is true when a1 = az and bi = bz. (c) From the table of mgfs we see that the mgfs of X and Y are Vx(t) = exp{(t – a,)* /2}, Vy (t) = exp{(t – az)² /2}. By the convolution property, Vz(t) = ¥x(t)¥y(t) = 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 Vx(t) = exp{t* /2}, Vy (t) = exp{t* /2}. By the convolution property, Vz(t) = ¥x(t)¥y (t) = exp{2t² /2}. This is an mgf of a normal random variable. (e) From the table of mgfs we see that the mgfs of X and Y are Vx(t) = exp{a,t + (b;t² /2)}, Vy (t) = exp{azt + (b;t² /2)}. By the convolution property, ¥z(t) = ¥x(t)¥y (t) = exp{(a1 + az)t + ((b} + b3)t² /2)}. This is an mgf of a N(a, + a2, (b + b;)) random variable.
A First Course in Probability (10th Edition)
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Chapter1: Combinatorial Analysis
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Transcribed Image Text:(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.
(c) The statement of the exercise is true when aj = az and b, = bz.
(c) From the table of mgfs we see that the mgfs of X and Y are
Vx(t) = exp{(t – a1) /2},
Vy (t) = exp{(t – az)² /2}.
By the convolution property,
Vz(t) = Vx(t)¥y (t) = 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
Vx(t) = exp{t° /2},
Vy (t) = exp{t² /2}.
By the convolution property,
Vz(t) = ¥x(t)¥y (t) = exp{2t² /2}.
This is an mgf of a normal random variable.
(e) From the table of mgfs we see that the mgfs of X and Y are
Vx(t) = exp{a,t+ (b;t² /2)},
Vy (t) = exp{azt + (b3t² /2)}.
By the convolution property,
Vz(t) = ¥x(t)¥y(t) = exp{(a, + az)t + ((b}+ b3)t² /2)}.
This is an mgf of a N(a, + az, (b + b3)) random variable.
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