1
$\begingroup$

-UPDATED SOLUTION from sasha's hint-

Thanks for checking my solution and see if i made any conceptual error.

the question http://dl.dropbox.com/u/5681270/Screen%20Shot%202011-11-09%20at%201.40.47%20PM.png

(i)

$f_{x}(x,y) = \int_{0}^{\infty} e^{-y} dy = 1 $ (uniform in 0 to 1)

$f_{y}(x,y) = \int_{0}^{1} e^{-y} dx = e^{-y} $ (exponential with unit mean)

from the above, $f(x,y) = f_{X}(x)f_{Y}(y)$, therefore they are independent

$F_v = P(min{X,Y}>v) = P(X>v)*P(Y>v) = (1-F_{x}(v))(1-F_{y}(v))$

$F_{X}(v) = v$ and $F_{Y}(v) = 1-e^{-v}$

therefore, $\frac{dF_{V}(v)}{dv}$

$ f_{min{X,Y}}(v) = \left\{ \begin{array}{lr} (v-2)e^{-v}-1 & : 00 \\ 0 & : otherwise \end{array} \right. $

(ii)

note that 2X has uniform distribution over $(o,2)$,

by convolution theorem,

$f_{2X+Y}(w) = \int_{-\infty}^{\infty}f_{2x}(x)*f_{Y}(w-x)dx$

$= \int_{0}^{2}\frac{1}{2}f_{Y}(w-x)dx$ since $f_{2x}(x) = \frac{1}{2}$ in $(0,2)$

as x goes from 0 to 2, y goes from w to w-2,

$= \frac{1}{2} \int_{w-2}^{w} f_{Y}(z)dz$

from here, we'll have 2 cases, (a) $0 < w < 2$ (b) $w > 2$

in (a)

$f_{2X + Y}(w) = \int_{0}^{w} e^(-z) dz = 1 - e^{-w}$

in (b)

$f_{2X + Y}(w) = \int_{w-2}^{w} e^(-z) dz = e^{-w - 2} - e^{-2}$

therefore, $ f_{2X + Y}(w) = \left\{ \begin{array}{lr} 1 - e^{-w} & : 0 < w < 2 \\ e^{-w - 2} - e^{-2} & : w > 2 \\ 0 & : otherwise \end{array} \right. $

Didier, did i get it right this time round?

  • 0
    Precisely: as a self-teaching student, you will be glad to see that this is not an example, this is a method.2011-11-09

1 Answers 1

1

HINT: $(X,Y)$ are independent, $X$ uniform, and $Y$ exponential with unit mean.

For i), compute $\mathbb{P}( \min(X,Y) > z) = \mathbb{P}(X > z \land Y >z) = \mathbb{P}(X > z) \mathbb{P}( Y >z)$. Probability density can then be obtained by differentiation.

For ii), notice that $2 X$ has uniform distribution over $(0,2)$. The probability density of the sum of random variables is the convolution of densities of summands: $ f_{2 X + Y}(z) = \int_{-\infty}^\infty f_{2 X}(x) f_Y(z-x) \mathrm{d} x = \int_{0}^2 f_{2 X}(x) f_Y(z-x) \mathrm{d} x $

  • 0
    the only thing i can think of is the Jacobian, is that right? how do i find it in such a case?2011-11-09