fX(x) = pfX|H(x) + (1-p)fX|T(x) = p + (1-p)2x
E[X] = p + (1-p)2/2 = p + (1-p)/2
P(Tails∣X=1/4) = (1-p)
a = p/(p+(1-p)2)
A random variable X is generated as follows. We flip a coin. With probability p , the result is Heads, and then X is generated according to a PDF fX|H which is uniform on [0,1] . With probability 1−p the result is Tails, and then X is generated according to a PDF fX|T of the form
fX|T(x)=2x,if x∈[0,1].
(The PDF is zero everywhere else.)
What is the (unconditional) PDF fX(x) of X ?
For 0≤x≤1 :
fX(x)=
Calculate E[X] =
Find P(Tails∣X=1/4) .
The MAP rule decides in favor of Heads if X<a and in favor of Tails if X>a. What is a?
a=
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