Finding $mathbbP(Y<g(X))$

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

Let $X$ and $Y$ be two independent random variables. The pdf of $Y$ is $f_Y(y)$ with $y geq a$, while the pdf of $X$ is

beginequation
f_X(x) =
begincases
f_X,1(x), quad b leq xleq c \
f_X,2(x), quad c < x leq d
endcases
endequation
where $a$, $b$, $c$ and $d$ are all positive real numbers such that $aleq b<c<d<infty$. Find $mathbbP[Y<g(X)]$ with $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.

Here is my latest attempt:
$$
mathbbP(Y<g(X)) =int_a^infty(1-F_g(X)(y)) f_Y(y) , dy
$$
Letting $Z=g(X)$.
beginalign*
F_Z(z)&=mathbbP(Z<z) \
&=mathbbP(g(X)<z)\
&=mathbbP(X<g^-1(z)) \
& = F_X(g^-1(z)) \
&=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*
Then,
beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

Now $g(.)$ is a monotonically increasing function. Hence, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the required probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)(1-F_X(g^-1(y)) f_Y(y) ,dy \
+ &int_g(b)^g(c)(1-F_X(g^-1(y)) f_Y(y) ,dy \
+ &int_g(c)^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(c)(1-F_X(g^-1(y)) f_Y(y) ,dy
+ int_g(c)^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

I am unsure how to proceed from here (or even if my attempt is correct to begin with). Any help with that?

probability probability-distributions

share|cite|improve this question

edited Jul 31 at 14:28

asked Jul 30 at 13:01

Lod

286

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You need to be more careful about the assumptions you're making as you go through every step. $g$ cannot be a "generic function." For starters, think about the following: what assumptions are you making when you go from $mathbbP(g(X) < z)$ to $mathbbP(X < g^-1(z))$? Hint: $g$ cannot be a "generic function" for this step to be valid.
  – Clarinetist
  Jul 30 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clarinetist In my case, the function is in the form $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.
  – Lod
  Jul 30 at 13:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus, $g(x)=ux^v$, for some positive $u$ and $v$? Anyway, without further information about the two PDFs involved, one can at most write down a general formula for $P(Y<g(X))$, not really compute it.
  – Did
  Jul 30 at 13:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did Yes that's true, and yes that's exactly what I would like to do: write down the general formula.
  – Lod
  Jul 30 at 13:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence, not a real question, as they say?
  – Did
  Jul 30 at 13:46
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

favorite

Let $X$ and $Y$ be two independent random variables. The pdf of $Y$ is $f_Y(y)$ with $y geq a$, while the pdf of $X$ is

beginequation
f_X(x) =
begincases
f_X,1(x), quad b leq xleq c \
f_X,2(x), quad c < x leq d
endcases
endequation
where $a$, $b$, $c$ and $d$ are all positive real numbers such that $aleq b<c<d<infty$. Find $mathbbP[Y<g(X)]$ with $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.

Here is my latest attempt:
$$
mathbbP(Y<g(X)) =int_a^infty(1-F_g(X)(y)) f_Y(y) , dy
$$
Letting $Z=g(X)$.
beginalign*
F_Z(z)&=mathbbP(Z<z) \
&=mathbbP(g(X)<z)\
&=mathbbP(X<g^-1(z)) \
& = F_X(g^-1(z)) \
&=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*
Then,
beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

Now $g(.)$ is a monotonically increasing function. Hence, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the required probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)(1-F_X(g^-1(y)) f_Y(y) ,dy \
+ &int_g(b)^g(c)(1-F_X(g^-1(y)) f_Y(y) ,dy \
+ &int_g(c)^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(c)(1-F_X(g^-1(y)) f_Y(y) ,dy
+ int_g(c)^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

I am unsure how to proceed from here (or even if my attempt is correct to begin with). Any help with that?

probability probability-distributions

share|cite|improve this question

edited Jul 31 at 14:28

asked Jul 30 at 13:01

Lod

286

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You need to be more careful about the assumptions you're making as you go through every step. $g$ cannot be a "generic function." For starters, think about the following: what assumptions are you making when you go from $mathbbP(g(X) < z)$ to $mathbbP(X < g^-1(z))$? Hint: $g$ cannot be a "generic function" for this step to be valid.
  – Clarinetist
  Jul 30 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clarinetist In my case, the function is in the form $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.
  – Lod
  Jul 30 at 13:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus, $g(x)=ux^v$, for some positive $u$ and $v$? Anyway, without further information about the two PDFs involved, one can at most write down a general formula for $P(Y<g(X))$, not really compute it.
  – Did
  Jul 30 at 13:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did Yes that's true, and yes that's exactly what I would like to do: write down the general formula.
  – Lod
  Jul 30 at 13:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence, not a real question, as they say?
  – Did
  Jul 30 at 13:46
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $X$ and $Y$ be two independent random variables. The pdf of $Y$ is $f_Y(y)$ with $y geq a$, while the pdf of $X$ is

beginequation
f_X(x) =
begincases
f_X,1(x), quad b leq xleq c \
f_X,2(x), quad c < x leq d
endcases
endequation
where $a$, $b$, $c$ and $d$ are all positive real numbers such that $aleq b<c<d<infty$. Find $mathbbP[Y<g(X)]$ with $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.

Here is my latest attempt:
$$
mathbbP(Y<g(X)) =int_a^infty(1-F_g(X)(y)) f_Y(y) , dy
$$
Letting $Z=g(X)$.
beginalign*
F_Z(z)&=mathbbP(Z<z) \
&=mathbbP(g(X)<z)\
&=mathbbP(X<g^-1(z)) \
& = F_X(g^-1(z)) \
&=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*
Then,
beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

Now $g(.)$ is a monotonically increasing function. Hence, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the required probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)(1-F_X(g^-1(y)) f_Y(y) ,dy \
+ &int_g(b)^g(c)(1-F_X(g^-1(y)) f_Y(y) ,dy \
+ &int_g(c)^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(c)(1-F_X(g^-1(y)) f_Y(y) ,dy
+ int_g(c)^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

I am unsure how to proceed from here (or even if my attempt is correct to begin with). Any help with that?

probability probability-distributions

share|cite|improve this question

edited Jul 31 at 14:28

asked Jul 30 at 13:01

Lod

286

Let $X$ and $Y$ be two independent random variables. The pdf of $Y$ is $f_Y(y)$ with $y geq a$, while the pdf of $X$ is

beginequation
f_X(x) =
begincases
f_X,1(x), quad b leq xleq c \
f_X,2(x), quad c < x leq d
endcases
endequation
where $a$, $b$, $c$ and $d$ are all positive real numbers such that $aleq b<c<d<infty$. Find $mathbbP[Y<g(X)]$ with $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.

Here is my latest attempt:
$$
mathbbP(Y<g(X)) =int_a^infty(1-F_g(X)(y)) f_Y(y) , dy
$$
Letting $Z=g(X)$.
beginalign*
F_Z(z)&=mathbbP(Z<z) \
&=mathbbP(g(X)<z)\
&=mathbbP(X<g^-1(z)) \
& = F_X(g^-1(z)) \
&=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*
Then,
beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

Now $g(.)$ is a monotonically increasing function. Hence, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the required probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)(1-F_X(g^-1(y)) f_Y(y) ,dy \
+ &int_g(b)^g(c)(1-F_X(g^-1(y)) f_Y(y) ,dy \
+ &int_g(c)^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(c)(1-F_X(g^-1(y)) f_Y(y) ,dy
+ int_g(c)^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

I am unsure how to proceed from here (or even if my attempt is correct to begin with). Any help with that?

probability probability-distributions

share|cite|improve this question

edited Jul 31 at 14:28

asked Jul 30 at 13:01

Lod

286

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 14:28

edited Jul 31 at 14:28

edited Jul 31 at 14:28

asked Jul 30 at 13:01

Lod

286

asked Jul 30 at 13:01

Lod

286

asked Jul 30 at 13:01

Lod

286

286

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You need to be more careful about the assumptions you're making as you go through every step. $g$ cannot be a "generic function." For starters, think about the following: what assumptions are you making when you go from $mathbbP(g(X) < z)$ to $mathbbP(X < g^-1(z))$? Hint: $g$ cannot be a "generic function" for this step to be valid.
  – Clarinetist
  Jul 30 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clarinetist In my case, the function is in the form $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.
  – Lod
  Jul 30 at 13:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus, $g(x)=ux^v$, for some positive $u$ and $v$? Anyway, without further information about the two PDFs involved, one can at most write down a general formula for $P(Y<g(X))$, not really compute it.
  – Did
  Jul 30 at 13:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did Yes that's true, and yes that's exactly what I would like to do: write down the general formula.
  – Lod
  Jul 30 at 13:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence, not a real question, as they say?
  – Did
  Jul 30 at 13:46
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You need to be more careful about the assumptions you're making as you go through every step. $g$ cannot be a "generic function." For starters, think about the following: what assumptions are you making when you go from $mathbbP(g(X) < z)$ to $mathbbP(X < g^-1(z))$? Hint: $g$ cannot be a "generic function" for this step to be valid.
  – Clarinetist
  Jul 30 at 13:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Clarinetist In my case, the function is in the form $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.
  – Lod
  Jul 30 at 13:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thus, $g(x)=ux^v$, for some positive $u$ and $v$? Anyway, without further information about the two PDFs involved, one can at most write down a general formula for $P(Y<g(X))$, not really compute it.
  – Did
  Jul 30 at 13:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Did Yes that's true, and yes that's exactly what I would like to do: write down the general formula.
  – Lod
  Jul 30 at 13:45
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hence, not a real question, as they say?
  – Did
  Jul 30 at 13:46
  

  
  
  
  

You need to be more careful about the assumptions you're making as you go through every step. $g$ cannot be a "generic function." For starters, think about the following: what assumptions are you making when you go from $mathbbP(g(X) < z)$ to $mathbbP(X < g^-1(z))$? Hint: $g$ cannot be a "generic function" for this step to be valid.
– Clarinetist
Jul 30 at 13:13

You need to be more careful about the assumptions you're making as you go through every step. $g$ cannot be a "generic function." For starters, think about the following: what assumptions are you making when you go from $mathbbP(g(X) < z)$ to $mathbbP(X < g^-1(z))$? Hint: $g$ cannot be a "generic function" for this step to be valid.
– Clarinetist
Jul 30 at 13:13

@Clarinetist In my case, the function is in the form $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.
– Lod
Jul 30 at 13:19

@Clarinetist In my case, the function is in the form $g(X)=(c_1 X^-c_2)^-1/c_3$ where $c_1$, $c_2$, and $c_3$ are positive real numbers.
– Lod
Jul 30 at 13:19

Thus, $g(x)=ux^v$, for some positive $u$ and $v$? Anyway, without further information about the two PDFs involved, one can at most write down a general formula for $P(Y<g(X))$, not really compute it.
– Did
Jul 30 at 13:42

Thus, $g(x)=ux^v$, for some positive $u$ and $v$? Anyway, without further information about the two PDFs involved, one can at most write down a general formula for $P(Y<g(X))$, not really compute it.
– Did
Jul 30 at 13:42

@Did Yes that's true, and yes that's exactly what I would like to do: write down the general formula.
– Lod
Jul 30 at 13:45

@Did Yes that's true, and yes that's exactly what I would like to do: write down the general formula.
– Lod
Jul 30 at 13:45

Hence, not a real question, as they say?
– Did
Jul 30 at 13:46

Hence, not a real question, as they say?
– Did
Jul 30 at 13:46

 | 
show 4 more comments

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

I'm posting this just in case someone else is tackling a problem similar to mine. I've come up with this solution after several trials and have verified it through numerical evaluation. I may be missing something though, given that I am a neophyte to probability.

beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

with
beginalign*
F_X(g^-1(z))
=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad quad quad quad quad quad quad , , , , g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*

Since $g(.)$ is a monotonically increasing function, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the desired probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)f_Y(y) ,dy \
+ &int_g(b)^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = &int_a^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

share|cite|improve this answer

answered Jul 31 at 16:53

Lod

286

add a comment | 

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accepted

I'm posting this just in case someone else is tackling a problem similar to mine. I've come up with this solution after several trials and have verified it through numerical evaluation. I may be missing something though, given that I am a neophyte to probability.

beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

with
beginalign*
F_X(g^-1(z))
=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad quad quad quad quad quad quad , , , , g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*

Since $g(.)$ is a monotonically increasing function, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the desired probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)f_Y(y) ,dy \
+ &int_g(b)^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = &int_a^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

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answered Jul 31 at 16:53

Lod

286

add a comment | 

up vote
0
down vote



accepted

I'm posting this just in case someone else is tackling a problem similar to mine. I've come up with this solution after several trials and have verified it through numerical evaluation. I may be missing something though, given that I am a neophyte to probability.

beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

with
beginalign*
F_X(g^-1(z))
=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad quad quad quad quad quad quad , , , , g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*

Since $g(.)$ is a monotonically increasing function, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the desired probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)f_Y(y) ,dy \
+ &int_g(b)^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = &int_a^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

share|cite|improve this answer

answered Jul 31 at 16:53

Lod

286

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

I'm posting this just in case someone else is tackling a problem similar to mine. I've come up with this solution after several trials and have verified it through numerical evaluation. I may be missing something though, given that I am a neophyte to probability.

beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

with
beginalign*
F_X(g^-1(z))
=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad quad quad quad quad quad quad , , , , g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*

Since $g(.)$ is a monotonically increasing function, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the desired probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)f_Y(y) ,dy \
+ &int_g(b)^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = &int_a^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

share|cite|improve this answer

answered Jul 31 at 16:53

Lod

286

I'm posting this just in case someone else is tackling a problem similar to mine. I've come up with this solution after several trials and have verified it through numerical evaluation. I may be missing something though, given that I am a neophyte to probability.

beginalign*
mathbbP(Y<g(X))= &int_a^infty (1-F_g(X)(y)) f_Y(y) , dy
= int_a^infty(1-F_X(g^-1(y)) f_Y(y) ,dy
endalign*

with
beginalign*
F_X(g^-1(z))
=
begincases
int_b^g^-1(z) f_X,1(x) ,dx, quad quad quad quad quad quad quad , , , , g(b) leq zleq g(c) \
int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dx, quad g(c) <z leq g(d)
endcases
endalign*

Since $g(.)$ is a monotonically increasing function, $g(b)<g(c)<g(d)$ and the solution depends on the value of $a$ relative to $g(b)$, $g(c)$ and $g(d)$. If $a>g(d)$, the desired probability is zero. Otherwise, we have three cases:

(1) If $a<g(b)$
beginalign*
mathbbP(Y<g(X)) = & int_a^g(b)f_Y(y) ,dy \
+ &int_g(b)^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(2) If $g(b)<a<g(c)$
beginalign*
mathbbP(Y<g(X)) = &int_a^g(c)left(1-int_b^g^-1(y) f_X,1(x) ,dxright) f_Y(y) ,dy \
+ &int_g(c)^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

(3) If $g(c)<a<g(d)$
beginalign*
mathbbP(Y<g(X)) = int_a^g(d)left(1- left[ int_b^c f_X,1(x) ,dx+int_c^g^-1(z) f_X,2(x) ,dxright]right) f_Y(y) ,dy
endalign*

share|cite|improve this answer

answered Jul 31 at 16:53

Lod

286

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 16:53

Lod

286

answered Jul 31 at 16:53

Lod

286

answered Jul 31 at 16:53

Lod

286

286

add a comment | 

add a comment | 

 

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