Find probability of $P(X Y < 1)$
$begingroup$
Let $X$ and $Y$ be two independent $mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$
I started off by finding the pdf
$f_X(x)=frac{1}{2} $
when $0<x<2$ Same for $Y$.
I then found their joint PDF via independence:
$f_{XY}(xy)=frac{1}{4} $ when $0leq y leq 2 $ and $0 leq x leq 2 $
Otherwise $0$
$int_0^2 int_0^{1/y} frac{1}{4} dx dy$
But this cannot be solved, so where did i go wrong?
probability probability-theory probability-distributions uniform-distribution
edited Nov 29 '18 at 9:58
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:24
WintherWinther
247
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be two independent $mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$
I started off by finding the pdf
$f_X(x)=frac{1}{2} $
when $0<x<2$ Same for $Y$.
I then found their joint PDF via independence:
$f_{XY}(xy)=frac{1}{4} $ when $0leq y leq 2 $ and $0 leq x leq 2 $
Otherwise $0$
$int_0^2 int_0^{1/y} frac{1}{4} dx dy$
But this cannot be solved, so where did i go wrong?
probability probability-theory probability-distributions uniform-distribution
edited Nov 29 '18 at 9:58
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:24
WintherWinther
247
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be two independent $mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$
I started off by finding the pdf
$f_X(x)=frac{1}{2} $
when $0<x<2$ Same for $Y$.
I then found their joint PDF via independence:
$f_{XY}(xy)=frac{1}{4} $ when $0leq y leq 2 $ and $0 leq x leq 2 $
Otherwise $0$
$int_0^2 int_0^{1/y} frac{1}{4} dx dy$
But this cannot be solved, so where did i go wrong?
probability probability-theory probability-distributions uniform-distribution
edited Nov 29 '18 at 9:58
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:24
WintherWinther
247
$endgroup$
Let $X$ and $Y$ be two independent $mathrm{Uniform}(0,2)$ random variables. Find $P(XY < 1)$
I started off by finding the pdf
$f_X(x)=frac{1}{2} $
when $0<x<2$ Same for $Y$.
I then found their joint PDF via independence:
$f_{XY}(xy)=frac{1}{4} $ when $0leq y leq 2 $ and $0 leq x leq 2 $
Otherwise $0$
$int_0^2 int_0^{1/y} frac{1}{4} dx dy$
But this cannot be solved, so where did i go wrong?
probability probability-theory probability-distributions uniform-distribution
probability probability-theory probability-distributions uniform-distribution
edited Nov 29 '18 at 9:58
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:24
WintherWinther
247
edited Nov 29 '18 at 9:58
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:24
WintherWinther
247
edited Nov 29 '18 at 9:58
Davide Giraudo
126k16150261
edited Nov 29 '18 at 9:58
Davide Giraudo
126k16150261
edited Nov 29 '18 at 9:58
Davide Giraudo
126k16150261
126k16150261
asked Nov 28 '18 at 21:24
WintherWinther
247
asked Nov 28 '18 at 21:24
WintherWinther
247
asked Nov 28 '18 at 21:24
WintherWinther
247
247
add a comment |
add a comment |
2 Answers
2
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votes
$begingroup$
Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).
$$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$
Your error was forgetting that the upper limit on $y$ was bounded by $2$.
answered Nov 28 '18 at 21:35
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
add a comment |
$begingroup$
Alternatively, you can consider the complementing blue area:
$hspace{2cm}$
$$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
&=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
&=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
&=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
&=frac14+frac12ln 2.end{align}$$
answered Nov 29 '18 at 12:36
farruhotafarruhota
20.1k2738
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).
$$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$
Your error was forgetting that the upper limit on $y$ was bounded by $2$.
answered Nov 28 '18 at 21:35
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
add a comment |
$begingroup$
Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).
$$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$
Your error was forgetting that the upper limit on $y$ was bounded by $2$.
answered Nov 28 '18 at 21:35
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
add a comment |
$begingroup$
Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).
$$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$
Your error was forgetting that the upper limit on $y$ was bounded by $2$.
answered Nov 28 '18 at 21:35
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
Here is the graph of $P(x,y)$ in the region of interest ($x y < 1$).
$$P[x y <1] = intlimits_{x=0}^2 intlimits_{y=0}^{min[2, 1/x]} {1 over 4} dy dx = frac{1}{4} (1+2 log (2))$$
Your error was forgetting that the upper limit on $y$ was bounded by $2$.
answered Nov 28 '18 at 21:35
David G. StorkDavid G. Stork
10.9k31432
edited Nov 28 '18 at 21:41
answered Nov 28 '18 at 21:35
David G. StorkDavid G. Stork
10.9k31432
answered Nov 28 '18 at 21:35
David G. StorkDavid G. Stork
10.9k31432
answered Nov 28 '18 at 21:35
David G. StorkDavid G. Stork
10.9k31432
10.9k31432
add a comment |
add a comment |
$begingroup$
Alternatively, you can consider the complementing blue area:
$hspace{2cm}$
$$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
&=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
&=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
&=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
&=frac14+frac12ln 2.end{align}$$
answered Nov 29 '18 at 12:36
farruhotafarruhota
20.1k2738
$endgroup$
add a comment |
$begingroup$
Alternatively, you can consider the complementing blue area:
$hspace{2cm}$
$$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
&=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
&=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
&=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
&=frac14+frac12ln 2.end{align}$$
answered Nov 29 '18 at 12:36
farruhotafarruhota
20.1k2738
$endgroup$
add a comment |
$begingroup$
Alternatively, you can consider the complementing blue area:
$hspace{2cm}$
$$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
&=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
&=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
&=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
&=frac14+frac12ln 2.end{align}$$
answered Nov 29 '18 at 12:36
farruhotafarruhota
20.1k2738
$endgroup$
Alternatively, you can consider the complementing blue area:
$hspace{2cm}$
$$begin{align}mathbb{P}(XY<1)&=1-mathbb{P}(XY>1)=\
&=1-int_{1/2}^2int_{1/x}^2 frac14dydx=\
&=1-int_{1/2}^2left(frac12-frac1{4x}right)dx=\
&=1-left(frac12x-frac14ln xright)big{|}_{1/2}^2=\
&=frac14+frac12ln 2.end{align}$$
answered Nov 29 '18 at 12:36
farruhotafarruhota
20.1k2738
answered Nov 29 '18 at 12:36
farruhotafarruhota
20.1k2738
answered Nov 29 '18 at 12:36
farruhotafarruhota
20.1k2738
answered Nov 29 '18 at 12:36
farruhotafarruhota
20.1k2738
20.1k2738
add a comment |
add a comment |
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