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Finding $mathbbP(P_S_1 , R_S_1^-alpha_S_1 > P_S_2 , R_S_2^-alpha_S_2)$
Clash Royale CLAN TAG#URR8PPP
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I have a system that consists of two subsystems and a user who is supposed to associate with either one of them. Let $S_1$ and $ S_2$ denote subsystem $1$ and subsystem $2$, respectively. The user associates with subsystem $k$ for $k=1,2$ if:
beginequation
P_S_k , R_S_k^-alpha_S_k > P_S_j , R_S_j^-alpha_S_j quad textfor all , , j=1,2 , ,textand , ,j ne k
endequation
where $P$ and $alpha$ are both positive real numbers that are preassigned to each subsystem (indicated by the index in the subscript) based on its features. On the other hand, $R_S_1$ and $R_S_2$ are two independent random variables. The pdf of $R_S_1$ is $f_R_S_1(r_s_1)$ with $r_s_1 geq a$, while the pdf of $R_S_2$ is a piece-wise function:
beginequation
f_R_S_2(r_s_2) =
begincases
f_R_S_2,1(r_s_2), quad b leq r_s_2 leq c \
f_R_S_2.2(r_s_2), quad c < r_s_2 leq d
endcases
endequation
where $a$, $b$, $c$ and $d$ are all positive real numbers such that $aleq b<c<d<infty$. Note that the support of $R_S_1$ is infinite, while the support of the $R_S_2$ is finite.
I want to find the association probability, denoted $mathcalA_k$ for $k=1,2$. It is the probability that the user is associated with $S_1$ or $S_2$. Here is my attempt for finding $mathcalA_1$ (Note that $mathcalA_2 = 1 -mathcalA_1$):
beginalign*
mathcalA_1&=mathbbP(P_S_1 , R_S_1^-alpha_S_1 > P_S_2 , R_S_2^-alpha_S_2)\
&=mathbbPleft( R_S_1 < left(fracP_S_2P_S_1 , R_S_2^-alpha_S_2right)^-frac1alpha_S_1right)\
&=mathbbPleft( R_S_1 < g(R_S_2) right)\
&= int_a^infty(1-F_g(R_S_2)(r_1)), f_R_S_1(r_1) ,mathrmdr_1 \
&= int_a^infty(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
&= int_a^g^-1(b)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1
+int_g^-1(b)^g^-1(c)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
&+int_g^-1(c)^g^-1(d)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1
+int_g^-1(d)^infty(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
endalign*
Now I need to substitute $F_R_S_2(g^-1(r_1))$ into the four integrals. In other words, I need to find $mathbbP(R_S_2 < g^-1(r_1))$ in each region, but I don't how. Any hints to get me started?
This question is an extension to the question posted here.
probability probability-distributions
asked Jul 25 at 8:47
Lod
336
add a comment |Â
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I have a system that consists of two subsystems and a user who is supposed to associate with either one of them. Let $S_1$ and $ S_2$ denote subsystem $1$ and subsystem $2$, respectively. The user associates with subsystem $k$ for $k=1,2$ if:
beginequation
P_S_k , R_S_k^-alpha_S_k > P_S_j , R_S_j^-alpha_S_j quad textfor all , , j=1,2 , ,textand , ,j ne k
endequation
where $P$ and $alpha$ are both positive real numbers that are preassigned to each subsystem (indicated by the index in the subscript) based on its features. On the other hand, $R_S_1$ and $R_S_2$ are two independent random variables. The pdf of $R_S_1$ is $f_R_S_1(r_s_1)$ with $r_s_1 geq a$, while the pdf of $R_S_2$ is a piece-wise function:
beginequation
f_R_S_2(r_s_2) =
begincases
f_R_S_2,1(r_s_2), quad b leq r_s_2 leq c \
f_R_S_2.2(r_s_2), quad c < r_s_2 leq d
endcases
endequation
where $a$, $b$, $c$ and $d$ are all positive real numbers such that $aleq b<c<d<infty$. Note that the support of $R_S_1$ is infinite, while the support of the $R_S_2$ is finite.
I want to find the association probability, denoted $mathcalA_k$ for $k=1,2$. It is the probability that the user is associated with $S_1$ or $S_2$. Here is my attempt for finding $mathcalA_1$ (Note that $mathcalA_2 = 1 -mathcalA_1$):
beginalign*
mathcalA_1&=mathbbP(P_S_1 , R_S_1^-alpha_S_1 > P_S_2 , R_S_2^-alpha_S_2)\
&=mathbbPleft( R_S_1 < left(fracP_S_2P_S_1 , R_S_2^-alpha_S_2right)^-frac1alpha_S_1right)\
&=mathbbPleft( R_S_1 < g(R_S_2) right)\
&= int_a^infty(1-F_g(R_S_2)(r_1)), f_R_S_1(r_1) ,mathrmdr_1 \
&= int_a^infty(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
&= int_a^g^-1(b)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1
+int_g^-1(b)^g^-1(c)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
&+int_g^-1(c)^g^-1(d)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1
+int_g^-1(d)^infty(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
endalign*
Now I need to substitute $F_R_S_2(g^-1(r_1))$ into the four integrals. In other words, I need to find $mathbbP(R_S_2 < g^-1(r_1))$ in each region, but I don't how. Any hints to get me started?
This question is an extension to the question posted here.
probability probability-distributions
asked Jul 25 at 8:47
Lod
336
1
A remark on the linked answer to your former question. Maybe I gave you the impression that conditional probabilities are indeed handsome or even necessary to find probabilities like P[X<Y]. I answered your former question along that lines in order to minimalize the deviation from the methods used in your question. But in my view a better way would have been the direct calculation of $mathbb E1_X<Y=int [x<y]f_X,Y(x,y)dydx$ where $[x<y]$ takes value $1$ if $x<y$ and takes value $0$ otherwise. So without any appeal on conditional probabilities.
– drhab
Jul 25 at 9:11
@drhab Thank you for the remark. I do get your point but still I'll need to divide in regions right? I mean I don't always know whether or not $X<Y$ (or in this case $R_S_1 < g(R_S_2)$. That's why I'm dividing the regions and conditioning on that. Pardon my ignorance. I'm an neophyte to probability and this to me is quite a complex problem.
– Lod
Jul 25 at 9:22
My remark only concerns the former question and I do not exclude that conditional probabilities can be handsome and/or necessary in solving this question (haven't looked at it enough). Good luck with probability.
– drhab
Jul 25 at 9:29
@drhab Oh in that case it makes sense. Thanks again!
– Lod
Jul 25 at 9:30
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I have a system that consists of two subsystems and a user who is supposed to associate with either one of them. Let $S_1$ and $ S_2$ denote subsystem $1$ and subsystem $2$, respectively. The user associates with subsystem $k$ for $k=1,2$ if:
beginequation
P_S_k , R_S_k^-alpha_S_k > P_S_j , R_S_j^-alpha_S_j quad textfor all , , j=1,2 , ,textand , ,j ne k
endequation
where $P$ and $alpha$ are both positive real numbers that are preassigned to each subsystem (indicated by the index in the subscript) based on its features. On the other hand, $R_S_1$ and $R_S_2$ are two independent random variables. The pdf of $R_S_1$ is $f_R_S_1(r_s_1)$ with $r_s_1 geq a$, while the pdf of $R_S_2$ is a piece-wise function:
beginequation
f_R_S_2(r_s_2) =
begincases
f_R_S_2,1(r_s_2), quad b leq r_s_2 leq c \
f_R_S_2.2(r_s_2), quad c < r_s_2 leq d
endcases
endequation
where $a$, $b$, $c$ and $d$ are all positive real numbers such that $aleq b<c<d<infty$. Note that the support of $R_S_1$ is infinite, while the support of the $R_S_2$ is finite.
I want to find the association probability, denoted $mathcalA_k$ for $k=1,2$. It is the probability that the user is associated with $S_1$ or $S_2$. Here is my attempt for finding $mathcalA_1$ (Note that $mathcalA_2 = 1 -mathcalA_1$):
beginalign*
mathcalA_1&=mathbbP(P_S_1 , R_S_1^-alpha_S_1 > P_S_2 , R_S_2^-alpha_S_2)\
&=mathbbPleft( R_S_1 < left(fracP_S_2P_S_1 , R_S_2^-alpha_S_2right)^-frac1alpha_S_1right)\
&=mathbbPleft( R_S_1 < g(R_S_2) right)\
&= int_a^infty(1-F_g(R_S_2)(r_1)), f_R_S_1(r_1) ,mathrmdr_1 \
&= int_a^infty(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
&= int_a^g^-1(b)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1
+int_g^-1(b)^g^-1(c)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
&+int_g^-1(c)^g^-1(d)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1
+int_g^-1(d)^infty(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
endalign*
Now I need to substitute $F_R_S_2(g^-1(r_1))$ into the four integrals. In other words, I need to find $mathbbP(R_S_2 < g^-1(r_1))$ in each region, but I don't how. Any hints to get me started?
This question is an extension to the question posted here.
probability probability-distributions
asked Jul 25 at 8:47
Lod
336
I have a system that consists of two subsystems and a user who is supposed to associate with either one of them. Let $S_1$ and $ S_2$ denote subsystem $1$ and subsystem $2$, respectively. The user associates with subsystem $k$ for $k=1,2$ if:
beginequation
P_S_k , R_S_k^-alpha_S_k > P_S_j , R_S_j^-alpha_S_j quad textfor all , , j=1,2 , ,textand , ,j ne k
endequation
where $P$ and $alpha$ are both positive real numbers that are preassigned to each subsystem (indicated by the index in the subscript) based on its features. On the other hand, $R_S_1$ and $R_S_2$ are two independent random variables. The pdf of $R_S_1$ is $f_R_S_1(r_s_1)$ with $r_s_1 geq a$, while the pdf of $R_S_2$ is a piece-wise function:
beginequation
f_R_S_2(r_s_2) =
begincases
f_R_S_2,1(r_s_2), quad b leq r_s_2 leq c \
f_R_S_2.2(r_s_2), quad c < r_s_2 leq d
endcases
endequation
where $a$, $b$, $c$ and $d$ are all positive real numbers such that $aleq b<c<d<infty$. Note that the support of $R_S_1$ is infinite, while the support of the $R_S_2$ is finite.
I want to find the association probability, denoted $mathcalA_k$ for $k=1,2$. It is the probability that the user is associated with $S_1$ or $S_2$. Here is my attempt for finding $mathcalA_1$ (Note that $mathcalA_2 = 1 -mathcalA_1$):
beginalign*
mathcalA_1&=mathbbP(P_S_1 , R_S_1^-alpha_S_1 > P_S_2 , R_S_2^-alpha_S_2)\
&=mathbbPleft( R_S_1 < left(fracP_S_2P_S_1 , R_S_2^-alpha_S_2right)^-frac1alpha_S_1right)\
&=mathbbPleft( R_S_1 < g(R_S_2) right)\
&= int_a^infty(1-F_g(R_S_2)(r_1)), f_R_S_1(r_1) ,mathrmdr_1 \
&= int_a^infty(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
&= int_a^g^-1(b)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1
+int_g^-1(b)^g^-1(c)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
&+int_g^-1(c)^g^-1(d)(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1
+int_g^-1(d)^infty(1-F_R_S_2(g^-1(r_1))), f_R_S_1(r_1) ,mathrmdr_1 \
endalign*
Now I need to substitute $F_R_S_2(g^-1(r_1))$ into the four integrals. In other words, I need to find $mathbbP(R_S_2 < g^-1(r_1))$ in each region, but I don't how. Any hints to get me started?
This question is an extension to the question posted here.
probability probability-distributions
asked Jul 25 at 8:47
Lod
336
edited Jul 30 at 12:12
asked Jul 25 at 8:47
Lod
336
asked Jul 25 at 8:47
Lod
336
asked Jul 25 at 8:47
Lod
336
336
1
A remark on the linked answer to your former question. Maybe I gave you the impression that conditional probabilities are indeed handsome or even necessary to find probabilities like P[X<Y]. I answered your former question along that lines in order to minimalize the deviation from the methods used in your question. But in my view a better way would have been the direct calculation of $mathbb E1_X<Y=int [x<y]f_X,Y(x,y)dydx$ where $[x<y]$ takes value $1$ if $x<y$ and takes value $0$ otherwise. So without any appeal on conditional probabilities.
– drhab
Jul 25 at 9:11
@drhab Thank you for the remark. I do get your point but still I'll need to divide in regions right? I mean I don't always know whether or not $X<Y$ (or in this case $R_S_1 < g(R_S_2)$. That's why I'm dividing the regions and conditioning on that. Pardon my ignorance. I'm an neophyte to probability and this to me is quite a complex problem.
– Lod
Jul 25 at 9:22
My remark only concerns the former question and I do not exclude that conditional probabilities can be handsome and/or necessary in solving this question (haven't looked at it enough). Good luck with probability.
– drhab
Jul 25 at 9:29
@drhab Oh in that case it makes sense. Thanks again!
– Lod
Jul 25 at 9:30
add a comment |Â
1
A remark on the linked answer to your former question. Maybe I gave you the impression that conditional probabilities are indeed handsome or even necessary to find probabilities like P[X<Y]. I answered your former question along that lines in order to minimalize the deviation from the methods used in your question. But in my view a better way would have been the direct calculation of $mathbb E1_X<Y=int [x<y]f_X,Y(x,y)dydx$ where $[x<y]$ takes value $1$ if $x<y$ and takes value $0$ otherwise. So without any appeal on conditional probabilities.
– drhab
Jul 25 at 9:11
@drhab Thank you for the remark. I do get your point but still I'll need to divide in regions right? I mean I don't always know whether or not $X<Y$ (or in this case $R_S_1 < g(R_S_2)$. That's why I'm dividing the regions and conditioning on that. Pardon my ignorance. I'm an neophyte to probability and this to me is quite a complex problem.
– Lod
Jul 25 at 9:22
My remark only concerns the former question and I do not exclude that conditional probabilities can be handsome and/or necessary in solving this question (haven't looked at it enough). Good luck with probability.
– drhab
Jul 25 at 9:29
@drhab Oh in that case it makes sense. Thanks again!
– Lod
Jul 25 at 9:30
1
1
A remark on the linked answer to your former question. Maybe I gave you the impression that conditional probabilities are indeed handsome or even necessary to find probabilities like P[X<Y]. I answered your former question along that lines in order to minimalize the deviation from the methods used in your question. But in my view a better way would have been the direct calculation of $mathbb E1_X<Y=int [x<y]f_X,Y(x,y)dydx$ where $[x<y]$ takes value $1$ if $x<y$ and takes value $0$ otherwise. So without any appeal on conditional probabilities.
– drhab
Jul 25 at 9:11
A remark on the linked answer to your former question. Maybe I gave you the impression that conditional probabilities are indeed handsome or even necessary to find probabilities like P[X<Y]. I answered your former question along that lines in order to minimalize the deviation from the methods used in your question. But in my view a better way would have been the direct calculation of $mathbb E1_X<Y=int [x<y]f_X,Y(x,y)dydx$ where $[x<y]$ takes value $1$ if $x<y$ and takes value $0$ otherwise. So without any appeal on conditional probabilities.
– drhab
Jul 25 at 9:11
@drhab Thank you for the remark. I do get your point but still I'll need to divide in regions right? I mean I don't always know whether or not $X<Y$ (or in this case $R_S_1 < g(R_S_2)$. That's why I'm dividing the regions and conditioning on that. Pardon my ignorance. I'm an neophyte to probability and this to me is quite a complex problem.
– Lod
Jul 25 at 9:22
@drhab Thank you for the remark. I do get your point but still I'll need to divide in regions right? I mean I don't always know whether or not $X<Y$ (or in this case $R_S_1 < g(R_S_2)$. That's why I'm dividing the regions and conditioning on that. Pardon my ignorance. I'm an neophyte to probability and this to me is quite a complex problem.
– Lod
Jul 25 at 9:22
My remark only concerns the former question and I do not exclude that conditional probabilities can be handsome and/or necessary in solving this question (haven't looked at it enough). Good luck with probability.
– drhab
Jul 25 at 9:29
My remark only concerns the former question and I do not exclude that conditional probabilities can be handsome and/or necessary in solving this question (haven't looked at it enough). Good luck with probability.
– drhab
Jul 25 at 9:29
@drhab Oh in that case it makes sense. Thanks again!
– Lod
Jul 25 at 9:30
@drhab Oh in that case it makes sense. Thanks again!
– Lod
Jul 25 at 9:30
add a comment |Â
1 Answer
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Unfortunately there is a mistake in the last expression.
Let $[x<y]$ denote the function that takes value $1$ if $x<y$ and value $0$ otherwise.
$$beginalignedmathbbPleft(R_S_1<gleft(R_S_2right)right) & =mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft[a,gleft(bright)right]right)+mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft(gleft(bright),gleft(dright)right]right)\
& =mathbbPleft(R_S_1inleft[a,gleft(bright)right]right)+int_gleft(bright)^gleft(dright)intleft[u<gleft(vright)right]f_R_S_1left(uright)f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)intleft[u<gleft(vright)right]f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_gleft(R_S_2right)left(uright)right)du\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_R_S_2left(g^-1left(uright)right)right)du
endaligned
$$
So in your last line $F_R_S_2,1(g(r_1))$ and $F_R_S_2,2(g(r_1))$ must
both be interchanged with $F_gleft(R_S_2right)(g^-1(r_1))$.
where $g^-1$ denotes the inverse function of the monotonically increasing function $g$.
Also observe that a random variable has only one CDF, so the notations $F_R_S_2,1$ and $F_R_S_2,2$ are not okay.
The split up in $left(gleft(bright),gleft(cright)right]$ and
$left(gleft(cright),gleft(dright)right]$ makes no sense.
answered Jul 25 at 11:29
drhab
86.2k541118
$F_gleft(R_S_2right)(u)$ is a CDF of a function of the random variable $R_S_2$—which will be $F_R_S_2left(left(fracP_S_1P_S_2 ,u^-alpha_S_1right)^-frac1alpha_S_2right)$. So I need to substitute the CDF of $R_S_2$, which in my case is going to be a piece-wise function, same as the pdf. That's why I have $F_R_S_2,1$ with $b leq r_s_2 leq c$ and $F_R_S_2,2$ with $c < r_s_2 leq d$, and that's why I divided the integrals. Now why wouldn't this be correct?
– Lod
Jul 25 at 12:14
I made a mistake by wrongly reading $F_R_S_2,1(r_1)$ (in stead of $F_R_S_2,1(g(r_1))$. Sorry, I will delete this within some minutes. Maybe things are okay after all, but I need a second look on it to confirm.
– drhab
Jul 25 at 12:27
Repaired now. There is still a mistake.
– drhab
Jul 25 at 12:35
Yeah exactly it should be the inverse. But what about you my replay to your remark regrading having only one CDF? I mean how else am I supposed to substitute $F_R_S_2$ given that it is piece-wise function?
– Lod
Jul 25 at 12:41
Notation $F_X$ stands for the CDF of rv $X$ which is a function on $mathbb R$.So writing $F_X,1$ and $F_X,2$ gives confusion. Also every function can be thought of as "piecewise function". The breaking up in pieces might become relevant by definition or calculation of integrals (e.g. when it concerns PDF's). This relevance lacks by CDF's. Actually I would rather not speak of piecewise functions, but of functions that are defined piecewise.
– drhab
Jul 25 at 12:49
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1 Answer
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Unfortunately there is a mistake in the last expression.
Let $[x<y]$ denote the function that takes value $1$ if $x<y$ and value $0$ otherwise.
$$beginalignedmathbbPleft(R_S_1<gleft(R_S_2right)right) & =mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft[a,gleft(bright)right]right)+mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft(gleft(bright),gleft(dright)right]right)\
& =mathbbPleft(R_S_1inleft[a,gleft(bright)right]right)+int_gleft(bright)^gleft(dright)intleft[u<gleft(vright)right]f_R_S_1left(uright)f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)intleft[u<gleft(vright)right]f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_gleft(R_S_2right)left(uright)right)du\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_R_S_2left(g^-1left(uright)right)right)du
endaligned
$$
So in your last line $F_R_S_2,1(g(r_1))$ and $F_R_S_2,2(g(r_1))$ must
both be interchanged with $F_gleft(R_S_2right)(g^-1(r_1))$.
where $g^-1$ denotes the inverse function of the monotonically increasing function $g$.
Also observe that a random variable has only one CDF, so the notations $F_R_S_2,1$ and $F_R_S_2,2$ are not okay.
The split up in $left(gleft(bright),gleft(cright)right]$ and
$left(gleft(cright),gleft(dright)right]$ makes no sense.
answered Jul 25 at 11:29
drhab
86.2k541118
$F_gleft(R_S_2right)(u)$ is a CDF of a function of the random variable $R_S_2$—which will be $F_R_S_2left(left(fracP_S_1P_S_2 ,u^-alpha_S_1right)^-frac1alpha_S_2right)$. So I need to substitute the CDF of $R_S_2$, which in my case is going to be a piece-wise function, same as the pdf. That's why I have $F_R_S_2,1$ with $b leq r_s_2 leq c$ and $F_R_S_2,2$ with $c < r_s_2 leq d$, and that's why I divided the integrals. Now why wouldn't this be correct?
– Lod
Jul 25 at 12:14
I made a mistake by wrongly reading $F_R_S_2,1(r_1)$ (in stead of $F_R_S_2,1(g(r_1))$. Sorry, I will delete this within some minutes. Maybe things are okay after all, but I need a second look on it to confirm.
– drhab
Jul 25 at 12:27
Repaired now. There is still a mistake.
– drhab
Jul 25 at 12:35
Yeah exactly it should be the inverse. But what about you my replay to your remark regrading having only one CDF? I mean how else am I supposed to substitute $F_R_S_2$ given that it is piece-wise function?
– Lod
Jul 25 at 12:41
Notation $F_X$ stands for the CDF of rv $X$ which is a function on $mathbb R$.So writing $F_X,1$ and $F_X,2$ gives confusion. Also every function can be thought of as "piecewise function". The breaking up in pieces might become relevant by definition or calculation of integrals (e.g. when it concerns PDF's). This relevance lacks by CDF's. Actually I would rather not speak of piecewise functions, but of functions that are defined piecewise.
– drhab
Jul 25 at 12:49
add a comment |Â
up vote
0
down vote
Unfortunately there is a mistake in the last expression.
Let $[x<y]$ denote the function that takes value $1$ if $x<y$ and value $0$ otherwise.
$$beginalignedmathbbPleft(R_S_1<gleft(R_S_2right)right) & =mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft[a,gleft(bright)right]right)+mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft(gleft(bright),gleft(dright)right]right)\
& =mathbbPleft(R_S_1inleft[a,gleft(bright)right]right)+int_gleft(bright)^gleft(dright)intleft[u<gleft(vright)right]f_R_S_1left(uright)f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)intleft[u<gleft(vright)right]f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_gleft(R_S_2right)left(uright)right)du\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_R_S_2left(g^-1left(uright)right)right)du
endaligned
$$
So in your last line $F_R_S_2,1(g(r_1))$ and $F_R_S_2,2(g(r_1))$ must
both be interchanged with $F_gleft(R_S_2right)(g^-1(r_1))$.
where $g^-1$ denotes the inverse function of the monotonically increasing function $g$.
Also observe that a random variable has only one CDF, so the notations $F_R_S_2,1$ and $F_R_S_2,2$ are not okay.
The split up in $left(gleft(bright),gleft(cright)right]$ and
$left(gleft(cright),gleft(dright)right]$ makes no sense.
answered Jul 25 at 11:29
drhab
86.2k541118
$F_gleft(R_S_2right)(u)$ is a CDF of a function of the random variable $R_S_2$—which will be $F_R_S_2left(left(fracP_S_1P_S_2 ,u^-alpha_S_1right)^-frac1alpha_S_2right)$. So I need to substitute the CDF of $R_S_2$, which in my case is going to be a piece-wise function, same as the pdf. That's why I have $F_R_S_2,1$ with $b leq r_s_2 leq c$ and $F_R_S_2,2$ with $c < r_s_2 leq d$, and that's why I divided the integrals. Now why wouldn't this be correct?
– Lod
Jul 25 at 12:14
I made a mistake by wrongly reading $F_R_S_2,1(r_1)$ (in stead of $F_R_S_2,1(g(r_1))$. Sorry, I will delete this within some minutes. Maybe things are okay after all, but I need a second look on it to confirm.
– drhab
Jul 25 at 12:27
Repaired now. There is still a mistake.
– drhab
Jul 25 at 12:35
Yeah exactly it should be the inverse. But what about you my replay to your remark regrading having only one CDF? I mean how else am I supposed to substitute $F_R_S_2$ given that it is piece-wise function?
– Lod
Jul 25 at 12:41
Notation $F_X$ stands for the CDF of rv $X$ which is a function on $mathbb R$.So writing $F_X,1$ and $F_X,2$ gives confusion. Also every function can be thought of as "piecewise function". The breaking up in pieces might become relevant by definition or calculation of integrals (e.g. when it concerns PDF's). This relevance lacks by CDF's. Actually I would rather not speak of piecewise functions, but of functions that are defined piecewise.
– drhab
Jul 25 at 12:49
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Unfortunately there is a mistake in the last expression.
Let $[x<y]$ denote the function that takes value $1$ if $x<y$ and value $0$ otherwise.
$$beginalignedmathbbPleft(R_S_1<gleft(R_S_2right)right) & =mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft[a,gleft(bright)right]right)+mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft(gleft(bright),gleft(dright)right]right)\
& =mathbbPleft(R_S_1inleft[a,gleft(bright)right]right)+int_gleft(bright)^gleft(dright)intleft[u<gleft(vright)right]f_R_S_1left(uright)f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)intleft[u<gleft(vright)right]f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_gleft(R_S_2right)left(uright)right)du\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_R_S_2left(g^-1left(uright)right)right)du
endaligned
$$
So in your last line $F_R_S_2,1(g(r_1))$ and $F_R_S_2,2(g(r_1))$ must
both be interchanged with $F_gleft(R_S_2right)(g^-1(r_1))$.
where $g^-1$ denotes the inverse function of the monotonically increasing function $g$.
Also observe that a random variable has only one CDF, so the notations $F_R_S_2,1$ and $F_R_S_2,2$ are not okay.
The split up in $left(gleft(bright),gleft(cright)right]$ and
$left(gleft(cright),gleft(dright)right]$ makes no sense.
answered Jul 25 at 11:29
drhab
86.2k541118
Unfortunately there is a mistake in the last expression.
Let $[x<y]$ denote the function that takes value $1$ if $x<y$ and value $0$ otherwise.
$$beginalignedmathbbPleft(R_S_1<gleft(R_S_2right)right) & =mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft[a,gleft(bright)right]right)+mathbbPleft(R_S_1<gleft(R_S_2right)wedge R_S_1inleft(gleft(bright),gleft(dright)right]right)\
& =mathbbPleft(R_S_1inleft[a,gleft(bright)right]right)+int_gleft(bright)^gleft(dright)intleft[u<gleft(vright)right]f_R_S_1left(uright)f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)intleft[u<gleft(vright)right]f_R_S_2left(vright)dvdu\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_gleft(R_S_2right)left(uright)right)du\
& =F_R_S_1left(gleft(bright)right)+int_gleft(bright)^gleft(dright)f_R_S_1left(uright)left(1-F_R_S_2left(g^-1left(uright)right)right)du
endaligned
$$
So in your last line $F_R_S_2,1(g(r_1))$ and $F_R_S_2,2(g(r_1))$ must
both be interchanged with $F_gleft(R_S_2right)(g^-1(r_1))$.
where $g^-1$ denotes the inverse function of the monotonically increasing function $g$.
Also observe that a random variable has only one CDF, so the notations $F_R_S_2,1$ and $F_R_S_2,2$ are not okay.
The split up in $left(gleft(bright),gleft(cright)right]$ and
$left(gleft(cright),gleft(dright)right]$ makes no sense.
answered Jul 25 at 11:29
drhab
86.2k541118
edited Jul 25 at 12:34
answered Jul 25 at 11:29
drhab
86.2k541118
answered Jul 25 at 11:29
drhab
86.2k541118
answered Jul 25 at 11:29
drhab
86.2k541118
86.2k541118
$F_gleft(R_S_2right)(u)$ is a CDF of a function of the random variable $R_S_2$—which will be $F_R_S_2left(left(fracP_S_1P_S_2 ,u^-alpha_S_1right)^-frac1alpha_S_2right)$. So I need to substitute the CDF of $R_S_2$, which in my case is going to be a piece-wise function, same as the pdf. That's why I have $F_R_S_2,1$ with $b leq r_s_2 leq c$ and $F_R_S_2,2$ with $c < r_s_2 leq d$, and that's why I divided the integrals. Now why wouldn't this be correct?
– Lod
Jul 25 at 12:14
I made a mistake by wrongly reading $F_R_S_2,1(r_1)$ (in stead of $F_R_S_2,1(g(r_1))$. Sorry, I will delete this within some minutes. Maybe things are okay after all, but I need a second look on it to confirm.
– drhab
Jul 25 at 12:27
Repaired now. There is still a mistake.
– drhab
Jul 25 at 12:35
Yeah exactly it should be the inverse. But what about you my replay to your remark regrading having only one CDF? I mean how else am I supposed to substitute $F_R_S_2$ given that it is piece-wise function?
– Lod
Jul 25 at 12:41
Notation $F_X$ stands for the CDF of rv $X$ which is a function on $mathbb R$.So writing $F_X,1$ and $F_X,2$ gives confusion. Also every function can be thought of as "piecewise function". The breaking up in pieces might become relevant by definition or calculation of integrals (e.g. when it concerns PDF's). This relevance lacks by CDF's. Actually I would rather not speak of piecewise functions, but of functions that are defined piecewise.
– drhab
Jul 25 at 12:49
add a comment |Â
$F_gleft(R_S_2right)(u)$ is a CDF of a function of the random variable $R_S_2$—which will be $F_R_S_2left(left(fracP_S_1P_S_2 ,u^-alpha_S_1right)^-frac1alpha_S_2right)$. So I need to substitute the CDF of $R_S_2$, which in my case is going to be a piece-wise function, same as the pdf. That's why I have $F_R_S_2,1$ with $b leq r_s_2 leq c$ and $F_R_S_2,2$ with $c < r_s_2 leq d$, and that's why I divided the integrals. Now why wouldn't this be correct?
– Lod
Jul 25 at 12:14
I made a mistake by wrongly reading $F_R_S_2,1(r_1)$ (in stead of $F_R_S_2,1(g(r_1))$. Sorry, I will delete this within some minutes. Maybe things are okay after all, but I need a second look on it to confirm.
– drhab
Jul 25 at 12:27
Repaired now. There is still a mistake.
– drhab
Jul 25 at 12:35
Yeah exactly it should be the inverse. But what about you my replay to your remark regrading having only one CDF? I mean how else am I supposed to substitute $F_R_S_2$ given that it is piece-wise function?
– Lod
Jul 25 at 12:41
Notation $F_X$ stands for the CDF of rv $X$ which is a function on $mathbb R$.So writing $F_X,1$ and $F_X,2$ gives confusion. Also every function can be thought of as "piecewise function". The breaking up in pieces might become relevant by definition or calculation of integrals (e.g. when it concerns PDF's). This relevance lacks by CDF's. Actually I would rather not speak of piecewise functions, but of functions that are defined piecewise.
– drhab
Jul 25 at 12:49
$F_gleft(R_S_2right)(u)$ is a CDF of a function of the random variable $R_S_2$—which will be $F_R_S_2left(left(fracP_S_1P_S_2 ,u^-alpha_S_1right)^-frac1alpha_S_2right)$. So I need to substitute the CDF of $R_S_2$, which in my case is going to be a piece-wise function, same as the pdf. That's why I have $F_R_S_2,1$ with $b leq r_s_2 leq c$ and $F_R_S_2,2$ with $c < r_s_2 leq d$, and that's why I divided the integrals. Now why wouldn't this be correct?
– Lod
Jul 25 at 12:14
$F_gleft(R_S_2right)(u)$ is a CDF of a function of the random variable $R_S_2$—which will be $F_R_S_2left(left(fracP_S_1P_S_2 ,u^-alpha_S_1right)^-frac1alpha_S_2right)$. So I need to substitute the CDF of $R_S_2$, which in my case is going to be a piece-wise function, same as the pdf. That's why I have $F_R_S_2,1$ with $b leq r_s_2 leq c$ and $F_R_S_2,2$ with $c < r_s_2 leq d$, and that's why I divided the integrals. Now why wouldn't this be correct?
– Lod
Jul 25 at 12:14
I made a mistake by wrongly reading $F_R_S_2,1(r_1)$ (in stead of $F_R_S_2,1(g(r_1))$. Sorry, I will delete this within some minutes. Maybe things are okay after all, but I need a second look on it to confirm.
– drhab
Jul 25 at 12:27
I made a mistake by wrongly reading $F_R_S_2,1(r_1)$ (in stead of $F_R_S_2,1(g(r_1))$. Sorry, I will delete this within some minutes. Maybe things are okay after all, but I need a second look on it to confirm.
– drhab
Jul 25 at 12:27
Repaired now. There is still a mistake.
– drhab
Jul 25 at 12:35
Repaired now. There is still a mistake.
– drhab
Jul 25 at 12:35
Yeah exactly it should be the inverse. But what about you my replay to your remark regrading having only one CDF? I mean how else am I supposed to substitute $F_R_S_2$ given that it is piece-wise function?
– Lod
Jul 25 at 12:41
Yeah exactly it should be the inverse. But what about you my replay to your remark regrading having only one CDF? I mean how else am I supposed to substitute $F_R_S_2$ given that it is piece-wise function?
– Lod
Jul 25 at 12:41
Notation $F_X$ stands for the CDF of rv $X$ which is a function on $mathbb R$.So writing $F_X,1$ and $F_X,2$ gives confusion. Also every function can be thought of as "piecewise function". The breaking up in pieces might become relevant by definition or calculation of integrals (e.g. when it concerns PDF's). This relevance lacks by CDF's. Actually I would rather not speak of piecewise functions, but of functions that are defined piecewise.
– drhab
Jul 25 at 12:49
Notation $F_X$ stands for the CDF of rv $X$ which is a function on $mathbb R$.So writing $F_X,1$ and $F_X,2$ gives confusion. Also every function can be thought of as "piecewise function". The breaking up in pieces might become relevant by definition or calculation of integrals (e.g. when it concerns PDF's). This relevance lacks by CDF's. Actually I would rather not speak of piecewise functions, but of functions that are defined piecewise.
– drhab
Jul 25 at 12:49
add a comment |Â
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1
A remark on the linked answer to your former question. Maybe I gave you the impression that conditional probabilities are indeed handsome or even necessary to find probabilities like P[X<Y]. I answered your former question along that lines in order to minimalize the deviation from the methods used in your question. But in my view a better way would have been the direct calculation of $mathbb E1_X<Y=int [x<y]f_X,Y(x,y)dydx$ where $[x<y]$ takes value $1$ if $x<y$ and takes value $0$ otherwise. So without any appeal on conditional probabilities.
– drhab
Jul 25 at 9:11
@drhab Thank you for the remark. I do get your point but still I'll need to divide in regions right? I mean I don't always know whether or not $X<Y$ (or in this case $R_S_1 < g(R_S_2)$. That's why I'm dividing the regions and conditioning on that. Pardon my ignorance. I'm an neophyte to probability and this to me is quite a complex problem.
– Lod
Jul 25 at 9:22
My remark only concerns the former question and I do not exclude that conditional probabilities can be handsome and/or necessary in solving this question (haven't looked at it enough). Good luck with probability.
– drhab
Jul 25 at 9:29
@drhab Oh in that case it makes sense. Thanks again!
– Lod
Jul 25 at 9:30