Question

3) Consider a random variable X which is uniformly distributed between [0,1] and the event A = {X > \}. a) Determine Hy = E{X}and o = E{(X - hy)?}. b) Determine the conditional distribution function f(x4) c) Determine E{XA}.

Answer 1

a) $p(X=x)=1\text{ for }x\in [0,1]$

Mean value is $\mu_X=\int_0^1 xp(x)\,dx=\int_0^1 x\,dx=\frac{1}{2}$

And $\sigma_X^2=E(X-\mu_X)^2=E(X-0.5)^2=\int_0^1(x-0.5)^2\,dx=\frac{1}{12}$

b) $f_x(x|A)=\frac{p(X=x)\cap p(x\in A)}{p(x\in A)}$

Where $p(X\in A)=\frac{1}{4}$ and $p((X=x)\cap (X\in A))=\frac{1}{4}\times 4=1$ for $x\in[3/4,1]$ and zero otherwise

So that $f_x(x|A)=\frac{1}{\frac{1}{4}}=4$ for $x\in[3/4,1]$ and zero otherwise

c) $E(X|A)=\int xf_x(x|A)\,dx=\int_{3/4}^{1} 4xdx=0.875$

Which is logical as we expect the middle value namely $\frac{\left(\frac{3}{4}+1\right)}{2}=0.875$ to be the mean of this new variable

$\blacksquare$

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