Let X and Y be two independent random variables with X ~ Poisson(µy). Poisson(ux) and Y (a) Show that the moment generating function (mgf) of X is Mx(t) = exp{µx(e' – 1)} = eHx(et –1). (b) Show that the mgf of X + Y is Mx+y(t) = exp{(ux + µy)(e' – 1)} (c) Based on the result in (b), what distribution does X +Y follow? (d) Use the mgf you found in (a) to show that E(X²) = µ + µx.
Let X and Y be two independent random variables with X ~ Poisson(µy). Poisson(ux) and Y (a) Show that the moment generating function (mgf) of X is Mx(t) = exp{µx(e' – 1)} = eHx(et –1). (b) Show that the mgf of X + Y is Mx+y(t) = exp{(ux + µy)(e' – 1)} (c) Based on the result in (b), what distribution does X +Y follow? (d) Use the mgf you found in (a) to show that E(X²) = µ + µx.
Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:HOUGHTON MIFFLIN HARCOURT
Chapter4: Writing Linear Equations
Section: Chapter Questions
Problem 12CR
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![Let X and Y be two independent random variables with X~ Poisson(ux) and Y ~
Poisson(µy).
(a) Show that the moment generating function (mgf) of X is Mx(t) = exp{µx (e² – 1)} =
eHx(et –1).
(b) Show that the mgf of X + Y is Mx+y(t) = exp{(µx + µy)(e* – 1)}
(c) Based on the result in (b), what distribution does X +Y follow?
(d) Use the mgf you found in (a) to show that E(X²) = µ3 + µx.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4db5c14e-f763-4823-bed1-193919508842%2F1cb3724e-32d1-49ec-a9ce-ebd3756d7d26%2Frlm3b5r_processed.png&w=3840&q=75)
Transcribed Image Text:Let X and Y be two independent random variables with X~ Poisson(ux) and Y ~
Poisson(µy).
(a) Show that the moment generating function (mgf) of X is Mx(t) = exp{µx (e² – 1)} =
eHx(et –1).
(b) Show that the mgf of X + Y is Mx+y(t) = exp{(µx + µy)(e* – 1)}
(c) Based on the result in (b), what distribution does X +Y follow?
(d) Use the mgf you found in (a) to show that E(X²) = µ3 + µx.
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