Mixing
Finding the cumulative distribution function of the sum of two functions of two independent random variables
$begingroup$
Since we want two real roots, then we're looking for a positive discriminant.
Let $Z$ be the random variable representing the value taken by the discriminant.
So,
$$Z = 4(U^2 - V)$$
However, I don't know where to take it from here.
Any Hints would be really appreciated.
Thanks
probability probability-distributions random-variables
asked Jan 5 at 2:53
SulSul
333114
$endgroup$
add a comment |
$begingroup$
Since we want two real roots, then we're looking for a positive discriminant.
Let $Z$ be the random variable representing the value taken by the discriminant.
So,
$$Z = 4(U^2 - V)$$
However, I don't know where to take it from here.
Any Hints would be really appreciated.
Thanks
probability probability-distributions random-variables
asked Jan 5 at 2:53
SulSul
333114
$endgroup$
add a comment |
$begingroup$
Since we want two real roots, then we're looking for a positive discriminant.
Let $Z$ be the random variable representing the value taken by the discriminant.
So,
$$Z = 4(U^2 - V)$$
However, I don't know where to take it from here.
Any Hints would be really appreciated.
Thanks
probability probability-distributions random-variables
asked Jan 5 at 2:53
SulSul
333114
$endgroup$
Since we want two real roots, then we're looking for a positive discriminant.
Let $Z$ be the random variable representing the value taken by the discriminant.
So,
$$Z = 4(U^2 - V)$$
However, I don't know where to take it from here.
Any Hints would be really appreciated.
Thanks
probability probability-distributions random-variables
probability probability-distributions random-variables
asked Jan 5 at 2:53
SulSul
333114
asked Jan 5 at 2:53
SulSul
333114
asked Jan 5 at 2:53
SulSul
333114
asked Jan 5 at 2:53
SulSul
333114
asked Jan 5 at 2:53
SulSul
333114
333114
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We want to find $mathbb P(Z>0)$, or $mathbb P(U^2>V)$. For $0<t<1$ we have
$$
mathbb P(U^2leqslant t) = mathbb P(Uleqslant t^{frac12}) = t^{frac12}
$$
and so $U^2$ has density
$$
f(t) = frac12 t^{-frac12}mathsf 1_{(0,1)}(t).
$$
We know already that $V$ has density $g(t) = mathsf 1_{(0,1)}(t)$, so integrating over the joint density we have
begin{align}
mathbb P(U^2>V) &= int_0^1 int_t^1 frac12 s^{-frac12} mathsf ds mathsf dt\
&= int_0^1 (1-t^{frac12}) mathsf dt\
&= frac13.
end{align}
answered Jan 5 at 3:18
Math1000Math1000
19k31745
$endgroup$
add a comment |
$begingroup$
$P{U^{2} >V}=EP({U^{2} >V}|U})=EU^{2}=int _0^{1} u^{2}, du =frac 1 3$.
answered Jan 5 at 5:09
Kavi Rama MurthyKavi Rama Murthy
54.8k32056
$endgroup$
add a comment |
$begingroup$
The positive discriminant implies $Z=4(U^2-V)>0$, hence you need to find $mathbb P(U^2-V>0)$. Refer to the graph:
$hspace{1cm}$
The required probability is the green area:
$$mathbb P(U^2-V>0)=mathbb P(V<U^2)=int_0^1 U^2 dU=frac{U^3}{3}bigg{|}_0^1=frac13.$$
answered Jan 5 at 6:31
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We want to find $mathbb P(Z>0)$, or $mathbb P(U^2>V)$. For $0<t<1$ we have
$$
mathbb P(U^2leqslant t) = mathbb P(Uleqslant t^{frac12}) = t^{frac12}
$$
and so $U^2$ has density
$$
f(t) = frac12 t^{-frac12}mathsf 1_{(0,1)}(t).
$$
We know already that $V$ has density $g(t) = mathsf 1_{(0,1)}(t)$, so integrating over the joint density we have
begin{align}
mathbb P(U^2>V) &= int_0^1 int_t^1 frac12 s^{-frac12} mathsf ds mathsf dt\
&= int_0^1 (1-t^{frac12}) mathsf dt\
&= frac13.
end{align}
answered Jan 5 at 3:18
Math1000Math1000
19k31745
$endgroup$
add a comment |
$begingroup$
We want to find $mathbb P(Z>0)$, or $mathbb P(U^2>V)$. For $0<t<1$ we have
$$
mathbb P(U^2leqslant t) = mathbb P(Uleqslant t^{frac12}) = t^{frac12}
$$
and so $U^2$ has density
$$
f(t) = frac12 t^{-frac12}mathsf 1_{(0,1)}(t).
$$
We know already that $V$ has density $g(t) = mathsf 1_{(0,1)}(t)$, so integrating over the joint density we have
begin{align}
mathbb P(U^2>V) &= int_0^1 int_t^1 frac12 s^{-frac12} mathsf ds mathsf dt\
&= int_0^1 (1-t^{frac12}) mathsf dt\
&= frac13.
end{align}
answered Jan 5 at 3:18
Math1000Math1000
19k31745
$endgroup$
add a comment |
$begingroup$
We want to find $mathbb P(Z>0)$, or $mathbb P(U^2>V)$. For $0<t<1$ we have
$$
mathbb P(U^2leqslant t) = mathbb P(Uleqslant t^{frac12}) = t^{frac12}
$$
and so $U^2$ has density
$$
f(t) = frac12 t^{-frac12}mathsf 1_{(0,1)}(t).
$$
We know already that $V$ has density $g(t) = mathsf 1_{(0,1)}(t)$, so integrating over the joint density we have
begin{align}
mathbb P(U^2>V) &= int_0^1 int_t^1 frac12 s^{-frac12} mathsf ds mathsf dt\
&= int_0^1 (1-t^{frac12}) mathsf dt\
&= frac13.
end{align}
answered Jan 5 at 3:18
Math1000Math1000
19k31745
$endgroup$
We want to find $mathbb P(Z>0)$, or $mathbb P(U^2>V)$. For $0<t<1$ we have
$$
mathbb P(U^2leqslant t) = mathbb P(Uleqslant t^{frac12}) = t^{frac12}
$$
and so $U^2$ has density
$$
f(t) = frac12 t^{-frac12}mathsf 1_{(0,1)}(t).
$$
We know already that $V$ has density $g(t) = mathsf 1_{(0,1)}(t)$, so integrating over the joint density we have
begin{align}
mathbb P(U^2>V) &= int_0^1 int_t^1 frac12 s^{-frac12} mathsf ds mathsf dt\
&= int_0^1 (1-t^{frac12}) mathsf dt\
&= frac13.
end{align}
answered Jan 5 at 3:18
Math1000Math1000
19k31745
answered Jan 5 at 3:18
Math1000Math1000
19k31745
answered Jan 5 at 3:18
Math1000Math1000
19k31745
answered Jan 5 at 3:18
Math1000Math1000
19k31745
19k31745
add a comment |
add a comment |
$begingroup$
$P{U^{2} >V}=EP({U^{2} >V}|U})=EU^{2}=int _0^{1} u^{2}, du =frac 1 3$.
answered Jan 5 at 5:09
Kavi Rama MurthyKavi Rama Murthy
54.8k32056
$endgroup$
add a comment |
$begingroup$
$P{U^{2} >V}=EP({U^{2} >V}|U})=EU^{2}=int _0^{1} u^{2}, du =frac 1 3$.
answered Jan 5 at 5:09
Kavi Rama MurthyKavi Rama Murthy
54.8k32056
$endgroup$
add a comment |
$begingroup$
$P{U^{2} >V}=EP({U^{2} >V}|U})=EU^{2}=int _0^{1} u^{2}, du =frac 1 3$.
answered Jan 5 at 5:09
Kavi Rama MurthyKavi Rama Murthy
54.8k32056
$endgroup$
$P{U^{2} >V}=EP({U^{2} >V}|U})=EU^{2}=int _0^{1} u^{2}, du =frac 1 3$.
answered Jan 5 at 5:09
Kavi Rama MurthyKavi Rama Murthy
54.8k32056
answered Jan 5 at 5:09
Kavi Rama MurthyKavi Rama Murthy
54.8k32056
answered Jan 5 at 5:09
Kavi Rama MurthyKavi Rama Murthy
54.8k32056
answered Jan 5 at 5:09
Kavi Rama MurthyKavi Rama Murthy
54.8k32056
54.8k32056
add a comment |
add a comment |
$begingroup$
The positive discriminant implies $Z=4(U^2-V)>0$, hence you need to find $mathbb P(U^2-V>0)$. Refer to the graph:
$hspace{1cm}$
The required probability is the green area:
$$mathbb P(U^2-V>0)=mathbb P(V<U^2)=int_0^1 U^2 dU=frac{U^3}{3}bigg{|}_0^1=frac13.$$
answered Jan 5 at 6:31
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
The positive discriminant implies $Z=4(U^2-V)>0$, hence you need to find $mathbb P(U^2-V>0)$. Refer to the graph:
$hspace{1cm}$
The required probability is the green area:
$$mathbb P(U^2-V>0)=mathbb P(V<U^2)=int_0^1 U^2 dU=frac{U^3}{3}bigg{|}_0^1=frac13.$$
answered Jan 5 at 6:31
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
The positive discriminant implies $Z=4(U^2-V)>0$, hence you need to find $mathbb P(U^2-V>0)$. Refer to the graph:
$hspace{1cm}$
The required probability is the green area:
$$mathbb P(U^2-V>0)=mathbb P(V<U^2)=int_0^1 U^2 dU=frac{U^3}{3}bigg{|}_0^1=frac13.$$
answered Jan 5 at 6:31
farruhotafarruhota
19.8k2738
$endgroup$
The positive discriminant implies $Z=4(U^2-V)>0$, hence you need to find $mathbb P(U^2-V>0)$. Refer to the graph:
$hspace{1cm}$
The required probability is the green area:
$$mathbb P(U^2-V>0)=mathbb P(V<U^2)=int_0^1 U^2 dU=frac{U^3}{3}bigg{|}_0^1=frac13.$$
answered Jan 5 at 6:31
farruhotafarruhota
19.8k2738
answered Jan 5 at 6:31
farruhotafarruhota
19.8k2738
answered Jan 5 at 6:31
farruhotafarruhota
19.8k2738
answered Jan 5 at 6:31
farruhotafarruhota
19.8k2738
19.8k2738
add a comment |
add a comment |
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