Mixing
How to drive the CDF of random variable $Z$ define by $ Z=frac{X_iX_j}{2X_i+X_j} $
$begingroup$
I am working in wireless communication and some times we use the PDFs and CDFs of random variables.
So I have read a paper and I found the derivation of CDF and PDF of random variable, but I did not understand the steps of the proof.
Let $X_1$ and $X_2$ two random variable with PDFs
$$
f_{X_i}(x)=frac{m_i^{m_i}}{Omega_i^{m_i}Gamma(m_i)}x^{m_i-1}e^{-frac{m_ix}{Omega_i}}.
$$
i want to find the CDF of random variable $Z$ define by
$$
Z=frac{X_iX_j}{2X_i+X_j}
$$
Now the athours of paper use the following steps:
begin{align}label{}
F_Z(z)=&mathbb{P}left{Zleq zright} \
=&mathbb{P}left{frac{X_iX_j}{2X_i+X_j}leq zright} \
stackrel{(a)}{=}&int_{0}^{infty}mathbb{P}left{frac{X_iX_j}{2X_i+X_j}leq z|X_iright}f_{X_i}(X_i)dX_i\
stackrel{(b)}{=}& int_{0}^{z}
mathbb{P}left{X_jgeqfrac{2zX_i}{X_i-z}|X_iright}
f_{X_i}(X_i)dX_i\
&+
int_{z}^{infty}
mathbb{P}left{X_jleqfrac{2zX_i}{X_i-z}|X_iright}
f_{X_i}(X_i)dX_i\
=&I_1(z)+I_2(z)
end{align}
where
$$
I_1(z)stackrel{(c)}{=}int_{0}^{z}f_{X_i}(X_i)dX_i
$$
$$
I_2(z)stackrel{(d)}{=}int_{z}^{infty}F_{X_j}(frac{2zX_i}{X_i-z})f_{X_i}(X_i)dX_i.
$$
My questions is are
Firs which low of probability is used in step (a)?.
Why hi divide integral in two range, from $0$ to $z$ and from $z$ to $infty$
Finally please some detail on steps (b), (c) and (d).
Thanks
probability probability-distributions conditional-probability
asked Jan 14 at 13:03
MonirMonir
278
$endgroup$
add a comment |
$begingroup$
I am working in wireless communication and some times we use the PDFs and CDFs of random variables.
So I have read a paper and I found the derivation of CDF and PDF of random variable, but I did not understand the steps of the proof.
Let $X_1$ and $X_2$ two random variable with PDFs
$$
f_{X_i}(x)=frac{m_i^{m_i}}{Omega_i^{m_i}Gamma(m_i)}x^{m_i-1}e^{-frac{m_ix}{Omega_i}}.
$$
i want to find the CDF of random variable $Z$ define by
$$
Z=frac{X_iX_j}{2X_i+X_j}
$$
Now the athours of paper use the following steps:
begin{align}label{}
F_Z(z)=&mathbb{P}left{Zleq zright} \
=&mathbb{P}left{frac{X_iX_j}{2X_i+X_j}leq zright} \
stackrel{(a)}{=}&int_{0}^{infty}mathbb{P}left{frac{X_iX_j}{2X_i+X_j}leq z|X_iright}f_{X_i}(X_i)dX_i\
stackrel{(b)}{=}& int_{0}^{z}
mathbb{P}left{X_jgeqfrac{2zX_i}{X_i-z}|X_iright}
f_{X_i}(X_i)dX_i\
&+
int_{z}^{infty}
mathbb{P}left{X_jleqfrac{2zX_i}{X_i-z}|X_iright}
f_{X_i}(X_i)dX_i\
=&I_1(z)+I_2(z)
end{align}
where
$$
I_1(z)stackrel{(c)}{=}int_{0}^{z}f_{X_i}(X_i)dX_i
$$
$$
I_2(z)stackrel{(d)}{=}int_{z}^{infty}F_{X_j}(frac{2zX_i}{X_i-z})f_{X_i}(X_i)dX_i.
$$
My questions is are
Firs which low of probability is used in step (a)?.
Why hi divide integral in two range, from $0$ to $z$ and from $z$ to $infty$
Finally please some detail on steps (b), (c) and (d).
Thanks
probability probability-distributions conditional-probability
asked Jan 14 at 13:03
MonirMonir
278
$endgroup$
1
$begingroup$
For step (a): the law of total probability. And for step (b), because the sign is different in the numerator $X_i-z$ depending on which side of $z$ you are, hence the inequality sign has to be carefully adapted as well.
$endgroup$
– Raskolnikov
Jan 14 at 13:12
$begingroup$
Hi, I did not understand this problem of sing?
$endgroup$
– Monir
Jan 14 at 13:16
add a comment |
$begingroup$
I am working in wireless communication and some times we use the PDFs and CDFs of random variables.
So I have read a paper and I found the derivation of CDF and PDF of random variable, but I did not understand the steps of the proof.
Let $X_1$ and $X_2$ two random variable with PDFs
$$
f_{X_i}(x)=frac{m_i^{m_i}}{Omega_i^{m_i}Gamma(m_i)}x^{m_i-1}e^{-frac{m_ix}{Omega_i}}.
$$
i want to find the CDF of random variable $Z$ define by
$$
Z=frac{X_iX_j}{2X_i+X_j}
$$
Now the athours of paper use the following steps:
begin{align}label{}
F_Z(z)=&mathbb{P}left{Zleq zright} \
=&mathbb{P}left{frac{X_iX_j}{2X_i+X_j}leq zright} \
stackrel{(a)}{=}&int_{0}^{infty}mathbb{P}left{frac{X_iX_j}{2X_i+X_j}leq z|X_iright}f_{X_i}(X_i)dX_i\
stackrel{(b)}{=}& int_{0}^{z}
mathbb{P}left{X_jgeqfrac{2zX_i}{X_i-z}|X_iright}
f_{X_i}(X_i)dX_i\
&+
int_{z}^{infty}
mathbb{P}left{X_jleqfrac{2zX_i}{X_i-z}|X_iright}
f_{X_i}(X_i)dX_i\
=&I_1(z)+I_2(z)
end{align}
where
$$
I_1(z)stackrel{(c)}{=}int_{0}^{z}f_{X_i}(X_i)dX_i
$$
$$
I_2(z)stackrel{(d)}{=}int_{z}^{infty}F_{X_j}(frac{2zX_i}{X_i-z})f_{X_i}(X_i)dX_i.
$$
My questions is are
Firs which low of probability is used in step (a)?.
Why hi divide integral in two range, from $0$ to $z$ and from $z$ to $infty$
Finally please some detail on steps (b), (c) and (d).
Thanks
probability probability-distributions conditional-probability
asked Jan 14 at 13:03
MonirMonir
278
$endgroup$
I am working in wireless communication and some times we use the PDFs and CDFs of random variables.
So I have read a paper and I found the derivation of CDF and PDF of random variable, but I did not understand the steps of the proof.
Let $X_1$ and $X_2$ two random variable with PDFs
$$
f_{X_i}(x)=frac{m_i^{m_i}}{Omega_i^{m_i}Gamma(m_i)}x^{m_i-1}e^{-frac{m_ix}{Omega_i}}.
$$
i want to find the CDF of random variable $Z$ define by
$$
Z=frac{X_iX_j}{2X_i+X_j}
$$
Now the athours of paper use the following steps:
begin{align}label{}
F_Z(z)=&mathbb{P}left{Zleq zright} \
=&mathbb{P}left{frac{X_iX_j}{2X_i+X_j}leq zright} \
stackrel{(a)}{=}&int_{0}^{infty}mathbb{P}left{frac{X_iX_j}{2X_i+X_j}leq z|X_iright}f_{X_i}(X_i)dX_i\
stackrel{(b)}{=}& int_{0}^{z}
mathbb{P}left{X_jgeqfrac{2zX_i}{X_i-z}|X_iright}
f_{X_i}(X_i)dX_i\
&+
int_{z}^{infty}
mathbb{P}left{X_jleqfrac{2zX_i}{X_i-z}|X_iright}
f_{X_i}(X_i)dX_i\
=&I_1(z)+I_2(z)
end{align}
where
$$
I_1(z)stackrel{(c)}{=}int_{0}^{z}f_{X_i}(X_i)dX_i
$$
$$
I_2(z)stackrel{(d)}{=}int_{z}^{infty}F_{X_j}(frac{2zX_i}{X_i-z})f_{X_i}(X_i)dX_i.
$$
My questions is are
Firs which low of probability is used in step (a)?.
Why hi divide integral in two range, from $0$ to $z$ and from $z$ to $infty$
Finally please some detail on steps (b), (c) and (d).
Thanks
probability probability-distributions conditional-probability
probability probability-distributions conditional-probability
asked Jan 14 at 13:03
MonirMonir
278
asked Jan 14 at 13:03
MonirMonir
278
asked Jan 14 at 13:03
MonirMonir
278
asked Jan 14 at 13:03
MonirMonir
278
asked Jan 14 at 13:03
MonirMonir
278
278
1
$begingroup$
For step (a): the law of total probability. And for step (b), because the sign is different in the numerator $X_i-z$ depending on which side of $z$ you are, hence the inequality sign has to be carefully adapted as well.
$endgroup$
– Raskolnikov
Jan 14 at 13:12
$begingroup$
Hi, I did not understand this problem of sing?
$endgroup$
– Monir
Jan 14 at 13:16
add a comment |
1
$begingroup$
For step (a): the law of total probability. And for step (b), because the sign is different in the numerator $X_i-z$ depending on which side of $z$ you are, hence the inequality sign has to be carefully adapted as well.
$endgroup$
– Raskolnikov
Jan 14 at 13:12
$begingroup$
Hi, I did not understand this problem of sing?
$endgroup$
– Monir
Jan 14 at 13:16
1
1
$begingroup$
For step (a): the law of total probability. And for step (b), because the sign is different in the numerator $X_i-z$ depending on which side of $z$ you are, hence the inequality sign has to be carefully adapted as well.
$endgroup$
– Raskolnikov
Jan 14 at 13:12
$begingroup$
For step (a): the law of total probability. And for step (b), because the sign is different in the numerator $X_i-z$ depending on which side of $z$ you are, hence the inequality sign has to be carefully adapted as well.
$endgroup$
– Raskolnikov
Jan 14 at 13:12
$begingroup$
Hi, I did not understand this problem of sing?
$endgroup$
– Monir
Jan 14 at 13:16
$begingroup$
Hi, I did not understand this problem of sing?
$endgroup$
– Monir
Jan 14 at 13:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A bit more detail on (b):
$$frac{X_iX_j}{2X_i+X_j}leq z Leftrightarrow X_iX_j leq 2zX_i+zX_j$$
and
$$X_iX_j leq 2zX_i+zX_j Leftrightarrow X_j(X_i-z) leq 2zX_i ; .$$
Now, in the last step, you have two possible cases: $X_i>z$ or $X_i < z$. If $X_i < z$ then the inequality is trivially true (because the $X_i$ are positive r.v.) and thus has probability 1. Hence why you end up with (c). The other case gives you (d).
answered Jan 14 at 14:43
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
$begingroup$
So you mean that $Pleft(X_jleq frac{2z}{X_i-z}right)=1$ for $X_ileq z$ otherwise Non?
$endgroup$
– Monir
Jan 14 at 14:51
$begingroup$
Why we flip $<$ to $>$ in interval $[0,z]$?
$endgroup$
– Monir
Jan 14 at 14:59
$begingroup$
Say we have $-3<-2$ and we want to carry over the $-2$. If you don't flip when carrying over the $-2$ you'd get $3/2 < 1$.
$endgroup$
– Raskolnikov
Jan 14 at 15:40
add a comment |
Your Answer
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1 Answer
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$begingroup$
A bit more detail on (b):
$$frac{X_iX_j}{2X_i+X_j}leq z Leftrightarrow X_iX_j leq 2zX_i+zX_j$$
and
$$X_iX_j leq 2zX_i+zX_j Leftrightarrow X_j(X_i-z) leq 2zX_i ; .$$
Now, in the last step, you have two possible cases: $X_i>z$ or $X_i < z$. If $X_i < z$ then the inequality is trivially true (because the $X_i$ are positive r.v.) and thus has probability 1. Hence why you end up with (c). The other case gives you (d).
answered Jan 14 at 14:43
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
$begingroup$
So you mean that $Pleft(X_jleq frac{2z}{X_i-z}right)=1$ for $X_ileq z$ otherwise Non?
$endgroup$
– Monir
Jan 14 at 14:51
$begingroup$
Why we flip $<$ to $>$ in interval $[0,z]$?
$endgroup$
– Monir
Jan 14 at 14:59
$begingroup$
Say we have $-3<-2$ and we want to carry over the $-2$. If you don't flip when carrying over the $-2$ you'd get $3/2 < 1$.
$endgroup$
– Raskolnikov
Jan 14 at 15:40
add a comment |
$begingroup$
A bit more detail on (b):
$$frac{X_iX_j}{2X_i+X_j}leq z Leftrightarrow X_iX_j leq 2zX_i+zX_j$$
and
$$X_iX_j leq 2zX_i+zX_j Leftrightarrow X_j(X_i-z) leq 2zX_i ; .$$
Now, in the last step, you have two possible cases: $X_i>z$ or $X_i < z$. If $X_i < z$ then the inequality is trivially true (because the $X_i$ are positive r.v.) and thus has probability 1. Hence why you end up with (c). The other case gives you (d).
answered Jan 14 at 14:43
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
$begingroup$
So you mean that $Pleft(X_jleq frac{2z}{X_i-z}right)=1$ for $X_ileq z$ otherwise Non?
$endgroup$
– Monir
Jan 14 at 14:51
$begingroup$
Why we flip $<$ to $>$ in interval $[0,z]$?
$endgroup$
– Monir
Jan 14 at 14:59
$begingroup$
Say we have $-3<-2$ and we want to carry over the $-2$. If you don't flip when carrying over the $-2$ you'd get $3/2 < 1$.
$endgroup$
– Raskolnikov
Jan 14 at 15:40
add a comment |
$begingroup$
A bit more detail on (b):
$$frac{X_iX_j}{2X_i+X_j}leq z Leftrightarrow X_iX_j leq 2zX_i+zX_j$$
and
$$X_iX_j leq 2zX_i+zX_j Leftrightarrow X_j(X_i-z) leq 2zX_i ; .$$
Now, in the last step, you have two possible cases: $X_i>z$ or $X_i < z$. If $X_i < z$ then the inequality is trivially true (because the $X_i$ are positive r.v.) and thus has probability 1. Hence why you end up with (c). The other case gives you (d).
answered Jan 14 at 14:43
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
A bit more detail on (b):
$$frac{X_iX_j}{2X_i+X_j}leq z Leftrightarrow X_iX_j leq 2zX_i+zX_j$$
and
$$X_iX_j leq 2zX_i+zX_j Leftrightarrow X_j(X_i-z) leq 2zX_i ; .$$
Now, in the last step, you have two possible cases: $X_i>z$ or $X_i < z$. If $X_i < z$ then the inequality is trivially true (because the $X_i$ are positive r.v.) and thus has probability 1. Hence why you end up with (c). The other case gives you (d).
answered Jan 14 at 14:43
RaskolnikovRaskolnikov
12.6k23571
answered Jan 14 at 14:43
RaskolnikovRaskolnikov
12.6k23571
answered Jan 14 at 14:43
RaskolnikovRaskolnikov
12.6k23571
answered Jan 14 at 14:43
RaskolnikovRaskolnikov
12.6k23571
12.6k23571
$begingroup$
So you mean that $Pleft(X_jleq frac{2z}{X_i-z}right)=1$ for $X_ileq z$ otherwise Non?
$endgroup$
– Monir
Jan 14 at 14:51
$begingroup$
Why we flip $<$ to $>$ in interval $[0,z]$?
$endgroup$
– Monir
Jan 14 at 14:59
$begingroup$
Say we have $-3<-2$ and we want to carry over the $-2$. If you don't flip when carrying over the $-2$ you'd get $3/2 < 1$.
$endgroup$
– Raskolnikov
Jan 14 at 15:40
add a comment |
$begingroup$
So you mean that $Pleft(X_jleq frac{2z}{X_i-z}right)=1$ for $X_ileq z$ otherwise Non?
$endgroup$
– Monir
Jan 14 at 14:51
$begingroup$
Why we flip $<$ to $>$ in interval $[0,z]$?
$endgroup$
– Monir
Jan 14 at 14:59
$begingroup$
Say we have $-3<-2$ and we want to carry over the $-2$. If you don't flip when carrying over the $-2$ you'd get $3/2 < 1$.
$endgroup$
– Raskolnikov
Jan 14 at 15:40
$begingroup$
So you mean that $Pleft(X_jleq frac{2z}{X_i-z}right)=1$ for $X_ileq z$ otherwise Non?
$endgroup$
– Monir
Jan 14 at 14:51
$begingroup$
So you mean that $Pleft(X_jleq frac{2z}{X_i-z}right)=1$ for $X_ileq z$ otherwise Non?
$endgroup$
– Monir
Jan 14 at 14:51
$begingroup$
Why we flip $<$ to $>$ in interval $[0,z]$?
$endgroup$
– Monir
Jan 14 at 14:59
$begingroup$
Why we flip $<$ to $>$ in interval $[0,z]$?
$endgroup$
– Monir
Jan 14 at 14:59
$begingroup$
Say we have $-3<-2$ and we want to carry over the $-2$. If you don't flip when carrying over the $-2$ you'd get $3/2 < 1$.
$endgroup$
– Raskolnikov
Jan 14 at 15:40
$begingroup$
Say we have $-3<-2$ and we want to carry over the $-2$. If you don't flip when carrying over the $-2$ you'd get $3/2 < 1$.
$endgroup$
– Raskolnikov
Jan 14 at 15:40
add a comment |
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1
$begingroup$
For step (a): the law of total probability. And for step (b), because the sign is different in the numerator $X_i-z$ depending on which side of $z$ you are, hence the inequality sign has to be carefully adapted as well.
$endgroup$
– Raskolnikov
Jan 14 at 13:12
$begingroup$
Hi, I did not understand this problem of sing?
$endgroup$
– Monir
Jan 14 at 13:16