Re: ISI 2007 ME-I - Doubts

Posted by Granpa Simpson on May 07, 2014; 6:09pm
URL: http://discussion-forum.276.s1.nabble.com/ISI-2007-ME-I-Doubts-tp7588851p7588926.html

Q14) Here, f(x;θ)=θ*f(x;1)+(1-θ)*f(x;0), where both f(x;1) and f(x;0) are pdf’s,
Now, ∫ f(x;θ)dx= θ*∫ f(x;1)dx+ (1-θ)*∫ f(x;0)dx.
Since ∫ f(x;1)dx=∫ f(x;0)dx=1.
∫ f(x;θ)dx=θ+(1-θ)=1.
Hence option d, f(x;θ) is a pdf for every θ

Q21) Integrate the function over (0,1), then (1,sqrt(2)), then (sqrt(2),3/2)) and you will get b as answer.