Question

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 f X|H which is uniform on [0,1] . With probability 1−p the result is Tails, and then X is generated according to a PDF f X|T of the form
f X|T (x)=2x,if x∈[0,1]. (The PDF is zero everywhere else.)
1. What is the (unconditional) PDF f X (x) of X ? For 0≤x≤1 : f X (x)=
2. Calculate E[X] .

Answer 1

Answer:

Following are the solution to the given points:

Step-by-step explanation:

For point a:

$fx|H(x) = 1;0< x<1\\\\fX|T(x) = 2x; 0\leq x \leq 1\\\\fx(x) = P(H \bigcap X = x) +P(T \bigcap X=x)$

$=P(H)fX|H(x)+P(T)fX|T(x)\\\\= p(1) + (1-p)2x\\\\= p(1 -2x)+2x$

Using the PDF of the X value

$fX(x) =2x +p(1 - 2x); \ 0\leq x\leq 1$

0 ; otherwise

For point b:

$E(X)=\int^{1}_{0} \ x fX (x)\ dx=\int^{1}_{0} \ x(2x+p(1-2x))\ dx\\\\=\int^{1}_{0} \ (2x^2+(x-2x^2)p) dx$

$= 2(\frac{x^3}{3}) + (\frac{x^2}{2}-2(\frac{x^3}{3}) \begin{vmatrix} x=1\\ x=0\end{vmatrix}$

$= \frac{2}{3} + (\frac{1}{2} - \frac{2}{3})p\\\\= \frac{2}{3} -\frac{p}{6}\\\\= \frac{(4 - p)}{6}$