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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


$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. $


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?

share|cite|improve this question
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 9 '11 at 6:51
@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 9 '11 at 8:25
Precisely: as a self-teaching student, you will be glad to see that this is not an example, this is a method. – Did Nov 9 '11 at 9:44
up vote 1 down vote accepted

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

share|cite|improve this answer
is my application of your hint right? – adsisco Nov 9 '11 at 9:37
@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 9 '11 at 13:14
$F_{Y}(v) = 1 - e^{-v}$? sorry but what is normalization factor? – adsisco Nov 9 '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 9 '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 9 '11 at 14:03

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