I do not know how to solve this exercise.. i have done some points of it by myself but i do not know if they are okay or not. This kind of exercises cost me a lot,my exam is coming soon :/
Can you help me to solve this exercise and make it easier for me to understand? It asks:
a)I have no idea :S
b)Mean proportion of impurities:
E(x) = ∫[0,1] x f(x) dx
= 2 ∫[0,1] x (1-x) dx
2 ∫ x (1-x) dx = 2 ( (1/2) x^2 - (1/3) x^3)
= x^2 -(2/3) x^3
F(x) = x^2-(2/3) x^3
F(1) = 1- 2/3 = 1/3
F(0) = 0
E(x) = 1/3-0 = 1/3 = 33.33 %
f(x) = 2-2x , 0 < x < 1
F(k) = ∫ [0,k] (2-2x) dx
= 2k-k^2
F(x) = 2x-x^2
How should i graph the cumulative distribution function of x?
c)
f(x) =2-2x , 0 < x < 1
find the density function of copper:
y=1-sqrt(x)
sqrt(x) = 1-y
x= (1-y)^2
2-2x= 2-2(1-y)^2
=2-2-2y^2+4y
=(4y-2y^2)
|dx/dy| = 2(1-y)(4y-2y^2) , 0 < y < 1 g(y) = (4y-2y^2) | dx/dy|
g(y) = 2(1-y)(4y-2y^2) , 0 < y < 1
g(y) = 4y^3-12y^2+8y , 0 < y < 1
P( y > 0.40) = ∫ [0.40,1] (4y^3-12y^2+8y) dy = 0.5904
Can you help me?
(b) Indeed $\mathsf E(X) ~{= \int_\Bbb R x f(x)\operatorname d x \\ = \int_0^1 x\cdot 2(1-x)\operatorname d x \\ = {[ x^2-\tfrac 23 x^3]}_{x=0}^{x=1} \\ = \tfrac 13 }$
Also the full expression for the CDF is: $F_X(x) = \begin{cases} 0 &:& x\lt 0 \\ 2x-x^2 &:& x\in (0;1) \\ 1 &:& x\in [1;\infty) \end{cases}$
(Noting the support is important)
And just plot that.
(c) Indeed, using the Jacobian transformation, $f_Y(y) = \begin{vmatrix}\dfrac{\mathrm d~x(y)}{\mathrm d~ y\quad~}\end{vmatrix} f_X(x(y))$ , where $x(y)=(1-y)^2$ for $0\leqslant y\leqslant 1$ we do obtain that:
$$f_Y(y) ~{=~ 2(1-y)\cdot 2(1-(1-y)^2)\cdot\mathbf 1_{y\in[0;1]} \\ = 4 y(y-1)(y-2)\cdot \mathbf 1_{y\in[0;1]} \\ = (4y^3-12y^2+8y)\cdot \mathbf 1_{y\in[0;1]} }$$
Then we indeed have: $1-F_Y(0.4) ~{= \int_{0.4}^1 (4y^3-12y^2+8y)\operatorname d y \\= 0.5904}$
We can confirm your answer to (c) without finding the denisty function of $Y$.
c) $\mathsf P(Y\gt 0.40) ~{= \mathsf P(\surd X\lt 0.60) \\ = \mathsf P(X\lt 0.36) \\ = \int_0^{0.36} 2(1-x)\operatorname d x \\ = 2\cdot 0.36-0.36^2 \\ = 0.5904}$
So now, you can do (a), similarly.
a) $\mathsf P(Y\leqslant 0.40\mid 0.20\leqslant X\leqslant 0.55) ~{ =\mathsf P(0.60\leqslant \surd X \mid 0.20\leqslant X\leqslant 0.55) \\ = \mathsf P(0.36\leqslant X\mid 0.20\leqslant X\leqslant 0.55) \\ = \dfrac{\mathsf P(0.36\leqslant X\leqslant 0.55)}{\mathsf P(0.20\leqslant X\leqslant 0.55)}\\ ~\vdots}$