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-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 & : 0<x<1, y>0 \\ 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?

adsisco
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  • Did you read this? This is the second time I mention this link to you and I think that if you had read it you would not make a mess of this exercise like you do above. Just my two cents. – Did Nov 09 '11 at 06:51
  • @Didier sorry, i'm self-teaching probability at the moment. so chances are, i'll often need more than an example to fully grasp the right concept, i'm trying to fix my solutions now, hopefully i'll get it right this time. – adsisco Nov 09 '11 at 08:25
  • Precisely: as a self-teaching student, you will be glad to see that this is not an example, this is a method. – Did Nov 09 '11 at 09:44

1 Answers1

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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: $$ f_{2 X + Y}(z) = \int_{-\infty}^\infty f_{2 X}(x) f_Y(z-x) \mathrm{d} x = \int_{0}^2 f_{2 X}(x) f_Y(z-x) \mathrm{d} x $$

Sasha
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  • is my application of your hint right? – adsisco Nov 09 '11 at 09:37
  • @adsisco $F_Y(v)$ in your post is incorrect. Remember, $F_Y(0) = \mathbb{P}(Y \le 0) = 0$, but the function you have gives 1. Also $\lim_{y \to \infty} F_Y(y) =0$, but your function does not have this property. Also for $f_{2X+Y}$ you forgot the normalization factor, and there is a typo in the formula. – Sasha Nov 09 '11 at 13:14
  • $F_{Y}(v) = 1 - e^{-v}$? sorry but what is normalization factor? – adsisco Nov 09 '11 at 13:56
  • @adsisco $\int_0^\infty f_{2X+Y}(w) \mathrm{d} w$ should equal to 1. Your current form of the function is such that this integral is not 1, but some other positive number. This can be fixed by multiplying your function by the reciprocal of this positive number, known as normalizing factor. – Sasha Nov 09 '11 at 13:59
  • the only thing i can think of is the Jacobian, is that right? how do i find it in such a case? – adsisco Nov 09 '11 at 14:03