density function of difference of two independent exponential random variables
$begingroup$
Let $X$ and $Y$ be independent exponential random variables with parameter 1. Find $P(Ygeq Xgeq 2)$.
My attempt:
Using independence of $X$ and $Y$ we get
$P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) = P(X-Yleq 0)(1-P(Xleq 2))=...=frac{1}{2}e^{-2}$
I used the fact that density of $X-Y$ is $f_{X-Y}(z)=frac{1}{2}e^{-|z|}$
However this turns out to be the wrong answer. Where did I go wrong?
probability
asked Nov 28 '18 at 6:01
dxdydzdxdydz
3009
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be independent exponential random variables with parameter 1. Find $P(Ygeq Xgeq 2)$.
My attempt:
Using independence of $X$ and $Y$ we get
$P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) = P(X-Yleq 0)(1-P(Xleq 2))=...=frac{1}{2}e^{-2}$
I used the fact that density of $X-Y$ is $f_{X-Y}(z)=frac{1}{2}e^{-|z|}$
However this turns out to be the wrong answer. Where did I go wrong?
probability
asked Nov 28 '18 at 6:01
dxdydzdxdydz
3009
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be independent exponential random variables with parameter 1. Find $P(Ygeq Xgeq 2)$.
My attempt:
Using independence of $X$ and $Y$ we get
$P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) = P(X-Yleq 0)(1-P(Xleq 2))=...=frac{1}{2}e^{-2}$
I used the fact that density of $X-Y$ is $f_{X-Y}(z)=frac{1}{2}e^{-|z|}$
However this turns out to be the wrong answer. Where did I go wrong?
probability
asked Nov 28 '18 at 6:01
dxdydzdxdydz
3009
$endgroup$
Let $X$ and $Y$ be independent exponential random variables with parameter 1. Find $P(Ygeq Xgeq 2)$.
My attempt:
Using independence of $X$ and $Y$ we get
$P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) = P(X-Yleq 0)(1-P(Xleq 2))=...=frac{1}{2}e^{-2}$
I used the fact that density of $X-Y$ is $f_{X-Y}(z)=frac{1}{2}e^{-|z|}$
However this turns out to be the wrong answer. Where did I go wrong?
probability
probability
asked Nov 28 '18 at 6:01
dxdydzdxdydz
3009
asked Nov 28 '18 at 6:01
dxdydzdxdydz
3009
asked Nov 28 '18 at 6:01
dxdydzdxdydz
3009
asked Nov 28 '18 at 6:01
dxdydzdxdydz
3009
asked Nov 28 '18 at 6:01
dxdydzdxdydz
3009
3009
add a comment |
add a comment |
1 Answer
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$begingroup$
Independence does not imply that $P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) $, since ${Yge X}$ and ${Xge2}$ are not necessarily independent events (just to start, they both are expressed in terms of $X$).
What independence tells you is that the joint density of $X$ and $Y$ is
$$f_{XY}(x,y)=f_X(x)cdot f_Y(y)=e^{-x}cdot e^{-y}$$
for $x,y>0$ (and $0$ in other case.)
So the probability you need is
$$iint_B e^{-(x+y)},dA,$$
where $B={(x,y)in mathbb R^2colon yge x ge 2, x>0, y>0}$. And this double integral is equivalent to the iterated integral
$$int_2^infty int_2^y e^{-(x+y)},dx,dy.$$
answered Nov 28 '18 at 6:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Independence does not imply that $P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) $, since ${Yge X}$ and ${Xge2}$ are not necessarily independent events (just to start, they both are expressed in terms of $X$).
What independence tells you is that the joint density of $X$ and $Y$ is
$$f_{XY}(x,y)=f_X(x)cdot f_Y(y)=e^{-x}cdot e^{-y}$$
for $x,y>0$ (and $0$ in other case.)
So the probability you need is
$$iint_B e^{-(x+y)},dA,$$
where $B={(x,y)in mathbb R^2colon yge x ge 2, x>0, y>0}$. And this double integral is equivalent to the iterated integral
$$int_2^infty int_2^y e^{-(x+y)},dx,dy.$$
answered Nov 28 '18 at 6:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
Independence does not imply that $P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) $, since ${Yge X}$ and ${Xge2}$ are not necessarily independent events (just to start, they both are expressed in terms of $X$).
What independence tells you is that the joint density of $X$ and $Y$ is
$$f_{XY}(x,y)=f_X(x)cdot f_Y(y)=e^{-x}cdot e^{-y}$$
for $x,y>0$ (and $0$ in other case.)
So the probability you need is
$$iint_B e^{-(x+y)},dA,$$
where $B={(x,y)in mathbb R^2colon yge x ge 2, x>0, y>0}$. And this double integral is equivalent to the iterated integral
$$int_2^infty int_2^y e^{-(x+y)},dx,dy.$$
answered Nov 28 '18 at 6:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
Independence does not imply that $P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) $, since ${Yge X}$ and ${Xge2}$ are not necessarily independent events (just to start, they both are expressed in terms of $X$).
What independence tells you is that the joint density of $X$ and $Y$ is
$$f_{XY}(x,y)=f_X(x)cdot f_Y(y)=e^{-x}cdot e^{-y}$$
for $x,y>0$ (and $0$ in other case.)
So the probability you need is
$$iint_B e^{-(x+y)},dA,$$
where $B={(x,y)in mathbb R^2colon yge x ge 2, x>0, y>0}$. And this double integral is equivalent to the iterated integral
$$int_2^infty int_2^y e^{-(x+y)},dx,dy.$$
answered Nov 28 '18 at 6:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
Independence does not imply that $P(Ygeq Xgeq 2) = P(Ygeq X) P(Xgeq 2) $, since ${Yge X}$ and ${Xge2}$ are not necessarily independent events (just to start, they both are expressed in terms of $X$).
What independence tells you is that the joint density of $X$ and $Y$ is
$$f_{XY}(x,y)=f_X(x)cdot f_Y(y)=e^{-x}cdot e^{-y}$$
for $x,y>0$ (and $0$ in other case.)
So the probability you need is
$$iint_B e^{-(x+y)},dA,$$
where $B={(x,y)in mathbb R^2colon yge x ge 2, x>0, y>0}$. And this double integral is equivalent to the iterated integral
$$int_2^infty int_2^y e^{-(x+y)},dx,dy.$$
answered Nov 28 '18 at 6:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Nov 28 '18 at 6:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Nov 28 '18 at 6:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Nov 28 '18 at 6:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
add a comment |
add a comment |
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