Mixing
Compute probability $P(U_1 leq atext{cos}(U_2))$
$begingroup$
Consider the independent random variables $U_1 =U(0,c)$ $U_2=U(0,pi/2)$
The challenge is to compute the probability:
$$P(U_1 leq atext{cos}(U_2))$$
An attempt:
$$P(U_1 leq atext{cos}(U_2))=P(U_1-acos(U_2)leq0)$$
Applying convolution:
$$int_{-infty}^0int_{-infty}^{infty} f_{U_1}(t)f_{-atext{cos}(U_2)}(x-t)dtdx$$
By independence:
$$=int_{-infty}^0int_{-infty}^{infty} f_{U_1,-atext{cos}(U_2)}(t,x-t)dtdx$$
This is where I get stuck, I believe using multivariate change of variable does the trick. The transform should give:
$$=int_0^{pi/2}int_0^{a text{cos(t)}} f_{U_1,U_2}(x,t)dxdt$$
From which it's no trouble to carry out the computation to the end. Is it correct to use multivariable change of variable in the above, and if so, what transformation should be performed?
calculus probability change-of-variable
asked Jan 21 at 6:43
DoleDole
926514
$endgroup$
|
show 4 more comments
$begingroup$
Consider the independent random variables $U_1 =U(0,c)$ $U_2=U(0,pi/2)$
The challenge is to compute the probability:
$$P(U_1 leq atext{cos}(U_2))$$
An attempt:
$$P(U_1 leq atext{cos}(U_2))=P(U_1-acos(U_2)leq0)$$
Applying convolution:
$$int_{-infty}^0int_{-infty}^{infty} f_{U_1}(t)f_{-atext{cos}(U_2)}(x-t)dtdx$$
By independence:
$$=int_{-infty}^0int_{-infty}^{infty} f_{U_1,-atext{cos}(U_2)}(t,x-t)dtdx$$
This is where I get stuck, I believe using multivariate change of variable does the trick. The transform should give:
$$=int_0^{pi/2}int_0^{a text{cos(t)}} f_{U_1,U_2}(x,t)dxdt$$
From which it's no trouble to carry out the computation to the end. Is it correct to use multivariable change of variable in the above, and if so, what transformation should be performed?
calculus probability change-of-variable
asked Jan 21 at 6:43
DoleDole
926514
$endgroup$
2
$begingroup$
begin{align} P(U_1le acos U_2)&=frac{2}{pi c}iint_{xle acos y}mathbf1_{0<x<c,0<y<pi/2},dx,dy \&=frac{2}{pi c}int_0^{pi/2}int_0^{min(c,acos y)},dx,dy end{align}
$endgroup$
– StubbornAtom
Jan 21 at 6:54
2
$begingroup$
You can just start directly with:$$P(U_1leq acos U_2)=mathbb Emathbf1_{U_1leq acos U_2}=intintmathbf1_{uleq acos v}f_{U,V}(u,v)dudv$$ leading to the expression mentioned by @StubbornAtom. No change of variables (or other tricks) is required.
$endgroup$
– drhab
Jan 21 at 9:54
$begingroup$
@drhab I can see that the method is quite favorable. But CAN this be solved using the transform?
$endgroup$
– Dole
Jan 22 at 0:03
$begingroup$
Why do you think that $P(U_1-acos U_2leq0)=P(U_1+acos U_2leq0)$?
$endgroup$
– drhab
Jan 22 at 10:11
$begingroup$
@drhab Good catch, corrected!
$endgroup$
– Dole
Jan 22 at 11:14
|
show 4 more comments
$begingroup$
Consider the independent random variables $U_1 =U(0,c)$ $U_2=U(0,pi/2)$
The challenge is to compute the probability:
$$P(U_1 leq atext{cos}(U_2))$$
An attempt:
$$P(U_1 leq atext{cos}(U_2))=P(U_1-acos(U_2)leq0)$$
Applying convolution:
$$int_{-infty}^0int_{-infty}^{infty} f_{U_1}(t)f_{-atext{cos}(U_2)}(x-t)dtdx$$
By independence:
$$=int_{-infty}^0int_{-infty}^{infty} f_{U_1,-atext{cos}(U_2)}(t,x-t)dtdx$$
This is where I get stuck, I believe using multivariate change of variable does the trick. The transform should give:
$$=int_0^{pi/2}int_0^{a text{cos(t)}} f_{U_1,U_2}(x,t)dxdt$$
From which it's no trouble to carry out the computation to the end. Is it correct to use multivariable change of variable in the above, and if so, what transformation should be performed?
calculus probability change-of-variable
asked Jan 21 at 6:43
DoleDole
926514
$endgroup$
Consider the independent random variables $U_1 =U(0,c)$ $U_2=U(0,pi/2)$
The challenge is to compute the probability:
$$P(U_1 leq atext{cos}(U_2))$$
An attempt:
$$P(U_1 leq atext{cos}(U_2))=P(U_1-acos(U_2)leq0)$$
Applying convolution:
$$int_{-infty}^0int_{-infty}^{infty} f_{U_1}(t)f_{-atext{cos}(U_2)}(x-t)dtdx$$
By independence:
$$=int_{-infty}^0int_{-infty}^{infty} f_{U_1,-atext{cos}(U_2)}(t,x-t)dtdx$$
This is where I get stuck, I believe using multivariate change of variable does the trick. The transform should give:
$$=int_0^{pi/2}int_0^{a text{cos(t)}} f_{U_1,U_2}(x,t)dxdt$$
From which it's no trouble to carry out the computation to the end. Is it correct to use multivariable change of variable in the above, and if so, what transformation should be performed?
calculus probability change-of-variable
calculus probability change-of-variable
asked Jan 21 at 6:43
DoleDole
926514
asked Jan 21 at 6:43
DoleDole
926514
edited Jan 22 at 13:10
Dole
asked Jan 21 at 6:43
DoleDole
926514
asked Jan 21 at 6:43
DoleDole
926514
asked Jan 21 at 6:43
DoleDole
926514
926514
2
$begingroup$
begin{align} P(U_1le acos U_2)&=frac{2}{pi c}iint_{xle acos y}mathbf1_{0<x<c,0<y<pi/2},dx,dy \&=frac{2}{pi c}int_0^{pi/2}int_0^{min(c,acos y)},dx,dy end{align}
$endgroup$
– StubbornAtom
Jan 21 at 6:54
2
$begingroup$
You can just start directly with:$$P(U_1leq acos U_2)=mathbb Emathbf1_{U_1leq acos U_2}=intintmathbf1_{uleq acos v}f_{U,V}(u,v)dudv$$ leading to the expression mentioned by @StubbornAtom. No change of variables (or other tricks) is required.
$endgroup$
– drhab
Jan 21 at 9:54
$begingroup$
@drhab I can see that the method is quite favorable. But CAN this be solved using the transform?
$endgroup$
– Dole
Jan 22 at 0:03
$begingroup$
Why do you think that $P(U_1-acos U_2leq0)=P(U_1+acos U_2leq0)$?
$endgroup$
– drhab
Jan 22 at 10:11
$begingroup$
@drhab Good catch, corrected!
$endgroup$
– Dole
Jan 22 at 11:14
|
show 4 more comments
2
$begingroup$
begin{align} P(U_1le acos U_2)&=frac{2}{pi c}iint_{xle acos y}mathbf1_{0<x<c,0<y<pi/2},dx,dy \&=frac{2}{pi c}int_0^{pi/2}int_0^{min(c,acos y)},dx,dy end{align}
$endgroup$
– StubbornAtom
Jan 21 at 6:54
2
$begingroup$
You can just start directly with:$$P(U_1leq acos U_2)=mathbb Emathbf1_{U_1leq acos U_2}=intintmathbf1_{uleq acos v}f_{U,V}(u,v)dudv$$ leading to the expression mentioned by @StubbornAtom. No change of variables (or other tricks) is required.
$endgroup$
– drhab
Jan 21 at 9:54
$begingroup$
@drhab I can see that the method is quite favorable. But CAN this be solved using the transform?
$endgroup$
– Dole
Jan 22 at 0:03
$begingroup$
Why do you think that $P(U_1-acos U_2leq0)=P(U_1+acos U_2leq0)$?
$endgroup$
– drhab
Jan 22 at 10:11
$begingroup$
@drhab Good catch, corrected!
$endgroup$
– Dole
Jan 22 at 11:14
2
2
$begingroup$
begin{align} P(U_1le acos U_2)&=frac{2}{pi c}iint_{xle acos y}mathbf1_{0<x<c,0<y<pi/2},dx,dy \&=frac{2}{pi c}int_0^{pi/2}int_0^{min(c,acos y)},dx,dy end{align}
$endgroup$
– StubbornAtom
Jan 21 at 6:54
$begingroup$
begin{align} P(U_1le acos U_2)&=frac{2}{pi c}iint_{xle acos y}mathbf1_{0<x<c,0<y<pi/2},dx,dy \&=frac{2}{pi c}int_0^{pi/2}int_0^{min(c,acos y)},dx,dy end{align}
$endgroup$
– StubbornAtom
Jan 21 at 6:54
2
2
$begingroup$
You can just start directly with:$$P(U_1leq acos U_2)=mathbb Emathbf1_{U_1leq acos U_2}=intintmathbf1_{uleq acos v}f_{U,V}(u,v)dudv$$ leading to the expression mentioned by @StubbornAtom. No change of variables (or other tricks) is required.
$endgroup$
– drhab
Jan 21 at 9:54
$begingroup$
You can just start directly with:$$P(U_1leq acos U_2)=mathbb Emathbf1_{U_1leq acos U_2}=intintmathbf1_{uleq acos v}f_{U,V}(u,v)dudv$$ leading to the expression mentioned by @StubbornAtom. No change of variables (or other tricks) is required.
$endgroup$
– drhab
Jan 21 at 9:54
$begingroup$
@drhab I can see that the method is quite favorable. But CAN this be solved using the transform?
$endgroup$
– Dole
Jan 22 at 0:03
$begingroup$
@drhab I can see that the method is quite favorable. But CAN this be solved using the transform?
$endgroup$
– Dole
Jan 22 at 0:03
$begingroup$
Why do you think that $P(U_1-acos U_2leq0)=P(U_1+acos U_2leq0)$?
$endgroup$
– drhab
Jan 22 at 10:11
$begingroup$
Why do you think that $P(U_1-acos U_2leq0)=P(U_1+acos U_2leq0)$?
$endgroup$
– drhab
Jan 22 at 10:11
$begingroup$
@drhab Good catch, corrected!
$endgroup$
– Dole
Jan 22 at 11:14
$begingroup$
@drhab Good catch, corrected!
$endgroup$
– Dole
Jan 22 at 11:14
|
show 4 more comments
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$begingroup$
begin{align} P(U_1le acos U_2)&=frac{2}{pi c}iint_{xle acos y}mathbf1_{0<x<c,0<y<pi/2},dx,dy \&=frac{2}{pi c}int_0^{pi/2}int_0^{min(c,acos y)},dx,dy end{align}
$endgroup$
– StubbornAtom
Jan 21 at 6:54
2
$begingroup$
You can just start directly with:$$P(U_1leq acos U_2)=mathbb Emathbf1_{U_1leq acos U_2}=intintmathbf1_{uleq acos v}f_{U,V}(u,v)dudv$$ leading to the expression mentioned by @StubbornAtom. No change of variables (or other tricks) is required.
$endgroup$
– drhab
Jan 21 at 9:54
$begingroup$
@drhab I can see that the method is quite favorable. But CAN this be solved using the transform?
$endgroup$
– Dole
Jan 22 at 0:03
$begingroup$
Why do you think that $P(U_1-acos U_2leq0)=P(U_1+acos U_2leq0)$?
$endgroup$
– drhab
Jan 22 at 10:11
$begingroup$
@drhab Good catch, corrected!
$endgroup$
– Dole
Jan 22 at 11:14