Mixing
Express moments in terms of exponential series
$begingroup$
Consider a random variable $X$ with cumulative distribution function
$$ F(x)= [G(x)]^{alpha} $$
where $ G(x)$ is baseline distribution, and its survival function is $ R_{G}(x)=1-G(x)$.
Put $R_{G}(x) $ in $F(x)$ it yields the density $f(x) = dF(x)/ d(x)$ as
$$ f(x)= alpha u'(x)e^{-u(x)}left(1 - e^{-u(x)} right)^{alpha-1}$$
where $u(x)= -ln R_G(x)$ and $u'(x)$ is the derivative $du(x) / dx$.
My question:
Show that the $rth$ moment of $X$ following the above pdf is given by
$$ E(X^{r})= r sum_{j=1}^{nu} c_{j} I_{j}(r) \
c_{j}= (-1)^{j} alpha (alpha-1)ldots (alpha-j+1),/,j! qquad I_{j}(r) = int_{0}^{infty} x^{r-1} e^{-j u(x)} dx $$
where $alpha > 0$ can be of non-integral value, and $nu in { alpha, infty }$.
integration sequences-and-series probability-distributions expected-value
edited Feb 2 at 13:25
Lee David Chung Lin
4,47841242
asked Jan 29 at 8:48
J.HJ.H
12
$endgroup$
|
show 2 more comments
$begingroup$
Consider a random variable $X$ with cumulative distribution function
$$ F(x)= [G(x)]^{alpha} $$
where $ G(x)$ is baseline distribution, and its survival function is $ R_{G}(x)=1-G(x)$.
Put $R_{G}(x) $ in $F(x)$ it yields the density $f(x) = dF(x)/ d(x)$ as
$$ f(x)= alpha u'(x)e^{-u(x)}left(1 - e^{-u(x)} right)^{alpha-1}$$
where $u(x)= -ln R_G(x)$ and $u'(x)$ is the derivative $du(x) / dx$.
My question:
Show that the $rth$ moment of $X$ following the above pdf is given by
$$ E(X^{r})= r sum_{j=1}^{nu} c_{j} I_{j}(r) \
c_{j}= (-1)^{j} alpha (alpha-1)ldots (alpha-j+1),/,j! qquad I_{j}(r) = int_{0}^{infty} x^{r-1} e^{-j u(x)} dx $$
where $alpha > 0$ can be of non-integral value, and $nu in { alpha, infty }$.
integration sequences-and-series probability-distributions expected-value
edited Feb 2 at 13:25
Lee David Chung Lin
4,47841242
asked Jan 29 at 8:48
J.HJ.H
12
$endgroup$
$begingroup$
@LeeDavidChungLin! yes [u(x)]'= du(x)/dx. It is derivative of u(x).
$endgroup$
– J.H
Jan 29 at 9:08
$begingroup$
$$ u(x)= -ln (R(x))$$ where R(x) is survival function. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 9:15
$begingroup$
Ok. Sorry for this. I do edit the question. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 11:24
$begingroup$
Yes I am still interested in answer. @LeeDavidChungLin
$endgroup$
– J.H
Feb 2 at 9:38
$begingroup$
@LeeDavidChungLin. our cdf is $$F(x) = (1-e^{-u(x)})^ {alpha} $$. I am also interested in expansion of its survival function (1-F(x)). which may be used to get moment.
$endgroup$
– J.H
Feb 2 at 9:52
|
show 2 more comments
$begingroup$
Consider a random variable $X$ with cumulative distribution function
$$ F(x)= [G(x)]^{alpha} $$
where $ G(x)$ is baseline distribution, and its survival function is $ R_{G}(x)=1-G(x)$.
Put $R_{G}(x) $ in $F(x)$ it yields the density $f(x) = dF(x)/ d(x)$ as
$$ f(x)= alpha u'(x)e^{-u(x)}left(1 - e^{-u(x)} right)^{alpha-1}$$
where $u(x)= -ln R_G(x)$ and $u'(x)$ is the derivative $du(x) / dx$.
My question:
Show that the $rth$ moment of $X$ following the above pdf is given by
$$ E(X^{r})= r sum_{j=1}^{nu} c_{j} I_{j}(r) \
c_{j}= (-1)^{j} alpha (alpha-1)ldots (alpha-j+1),/,j! qquad I_{j}(r) = int_{0}^{infty} x^{r-1} e^{-j u(x)} dx $$
where $alpha > 0$ can be of non-integral value, and $nu in { alpha, infty }$.
integration sequences-and-series probability-distributions expected-value
edited Feb 2 at 13:25
Lee David Chung Lin
4,47841242
asked Jan 29 at 8:48
J.HJ.H
12
$endgroup$
Consider a random variable $X$ with cumulative distribution function
$$ F(x)= [G(x)]^{alpha} $$
where $ G(x)$ is baseline distribution, and its survival function is $ R_{G}(x)=1-G(x)$.
Put $R_{G}(x) $ in $F(x)$ it yields the density $f(x) = dF(x)/ d(x)$ as
$$ f(x)= alpha u'(x)e^{-u(x)}left(1 - e^{-u(x)} right)^{alpha-1}$$
where $u(x)= -ln R_G(x)$ and $u'(x)$ is the derivative $du(x) / dx$.
My question:
Show that the $rth$ moment of $X$ following the above pdf is given by
$$ E(X^{r})= r sum_{j=1}^{nu} c_{j} I_{j}(r) \
c_{j}= (-1)^{j} alpha (alpha-1)ldots (alpha-j+1),/,j! qquad I_{j}(r) = int_{0}^{infty} x^{r-1} e^{-j u(x)} dx $$
where $alpha > 0$ can be of non-integral value, and $nu in { alpha, infty }$.
integration sequences-and-series probability-distributions expected-value
integration sequences-and-series probability-distributions expected-value
edited Feb 2 at 13:25
Lee David Chung Lin
4,47841242
asked Jan 29 at 8:48
J.HJ.H
12
edited Feb 2 at 13:25
Lee David Chung Lin
4,47841242
asked Jan 29 at 8:48
J.HJ.H
12
edited Feb 2 at 13:25
Lee David Chung Lin
4,47841242
edited Feb 2 at 13:25
Lee David Chung Lin
4,47841242
edited Feb 2 at 13:25
Lee David Chung Lin
4,47841242
4,47841242
asked Jan 29 at 8:48
J.HJ.H
12
asked Jan 29 at 8:48
J.HJ.H
12
asked Jan 29 at 8:48
J.HJ.H
12
12
$begingroup$
@LeeDavidChungLin! yes [u(x)]'= du(x)/dx. It is derivative of u(x).
$endgroup$
– J.H
Jan 29 at 9:08
$begingroup$
$$ u(x)= -ln (R(x))$$ where R(x) is survival function. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 9:15
$begingroup$
Ok. Sorry for this. I do edit the question. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 11:24
$begingroup$
Yes I am still interested in answer. @LeeDavidChungLin
$endgroup$
– J.H
Feb 2 at 9:38
$begingroup$
@LeeDavidChungLin. our cdf is $$F(x) = (1-e^{-u(x)})^ {alpha} $$. I am also interested in expansion of its survival function (1-F(x)). which may be used to get moment.
$endgroup$
– J.H
Feb 2 at 9:52
|
show 2 more comments
$begingroup$
@LeeDavidChungLin! yes [u(x)]'= du(x)/dx. It is derivative of u(x).
$endgroup$
– J.H
Jan 29 at 9:08
$begingroup$
$$ u(x)= -ln (R(x))$$ where R(x) is survival function. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 9:15
$begingroup$
Ok. Sorry for this. I do edit the question. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 11:24
$begingroup$
Yes I am still interested in answer. @LeeDavidChungLin
$endgroup$
– J.H
Feb 2 at 9:38
$begingroup$
@LeeDavidChungLin. our cdf is $$F(x) = (1-e^{-u(x)})^ {alpha} $$. I am also interested in expansion of its survival function (1-F(x)). which may be used to get moment.
$endgroup$
– J.H
Feb 2 at 9:52
$begingroup$
@LeeDavidChungLin! yes [u(x)]'= du(x)/dx. It is derivative of u(x).
$endgroup$
– J.H
Jan 29 at 9:08
$begingroup$
@LeeDavidChungLin! yes [u(x)]'= du(x)/dx. It is derivative of u(x).
$endgroup$
– J.H
Jan 29 at 9:08
$begingroup$
$$ u(x)= -ln (R(x))$$ where R(x) is survival function. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 9:15
$begingroup$
$$ u(x)= -ln (R(x))$$ where R(x) is survival function. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 9:15
$begingroup$
Ok. Sorry for this. I do edit the question. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 11:24
$begingroup$
Ok. Sorry for this. I do edit the question. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 11:24
$begingroup$
Yes I am still interested in answer. @LeeDavidChungLin
$endgroup$
– J.H
Feb 2 at 9:38
$begingroup$
Yes I am still interested in answer. @LeeDavidChungLin
$endgroup$
– J.H
Feb 2 at 9:38
$begingroup$
@LeeDavidChungLin. our cdf is $$F(x) = (1-e^{-u(x)})^ {alpha} $$. I am also interested in expansion of its survival function (1-F(x)). which may be used to get moment.
$endgroup$
– J.H
Feb 2 at 9:52
$begingroup$
@LeeDavidChungLin. our cdf is $$F(x) = (1-e^{-u(x)})^ {alpha} $$. I am also interested in expansion of its survival function (1-F(x)). which may be used to get moment.
$endgroup$
– J.H
Feb 2 at 9:52
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Please load the page twice for the hyperlinks to work properly
$require{begingroup}begingrouprenewcommand{dd}[1]{,mathrm{d}#1}$Allow me to reiterate the setting:
$G$ is the baseline CDF for a non-negative random variable.
$R_G equiv 1 - G$ as the baseline survival function.- Define $u equiv -log R_G$, or equivalently $R_G = e^{-u}$
- Consider a random variable $X$ which CDF is $F = G^alpha$, where $alpha > 0 $ can be non-integer.
- When $alpha in mathbb{N}$ is an integer, then $X overset{d}{=}max { W_1,~W_2,~ldots~, W_{alpha}}$, namely, $X$ has the same distribution as the max of an iid set with $G$ being their common CDF. At any rate, $F$ is well-defined for any non-integer $alpha > 0$.
- Taking $G = 1 - R_G = 1 - e^{-u}$ yields $F = (1 - e^{-u})^{alpha}$, which upon differentiating yields the density $f$ as shown in the question post.
The $r$-th moment of $X$ by the simplest definition is
$$E[X^r] = int_0^{infty} x^r f(x) dd{x}$$
Meanwhile, there's a well-known relation between the expectation and the integral of survival function. In particular, see the second half of this answer and another anwser from the statistics site for a visualization. In the current notation, it is
$$ E[X^r] = r int_0^{infty} x^{r-1} R_F(x) dd{x} $$
where $R_F equiv 1 - F$ is the survival function of $X$ such that after invoking the given item #6 in the beginning, we have
$$R_F(x) = 1 - left(1 - e^{-u(x)}right)^{alpha} label{eq_R_G_before_expansion} tag*{Eq.(1)}$$
Now, apply the series expansion (generalized Binomial theorem) on this quantity $1 - e^{-u(x)}$, which magnitude is smaller than unity, raised to the $alpha$-th power.
begin{align}
left(1 - e^{-u(x)}right)^{alpha} &= sum_{k = 0}^{infty} { alpha choose k} left(- e^{-u(x)} right)^k \
&= sum_{k = 0}^{infty} (-1)^k frac{ alpha! }{k! (alpha - k)! } e^{-k,u(x)} \
&= 1 - alpha e^{-u(x)} + frac{ alpha (alpha - 1) }{ 2! } e^{-2u(x)} - frac{ alpha (alpha - 1)(alpha - 2) }{ 3! } e^{-3u(x)} + cdots \
&= 1 + sum_{j = 1}^{infty~~text{or}~~alpha} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} label{eq_expansion_for_R_G} tag*{Eq.(2)}
end{align}
The leading $1$ is pulled out in anticipation of the next step, and the shift in the lower summation limit automatically accommodates the expression $(alpha - j + 1)$.
There are two cases for the upper summation limit:
- when $alpha$ is non-integer, the upper-limit is $infty$ as the series goes on forever.
- When $alpha in mathbb{N}$, the series is just the ordinary binomial expansion that terminates at $j = alpha$ with a total of $alpha + 1$ terms.
One can succinctly write the summation as $displaystyle sum_{j = 1}^{nu}~$ , where $nu in {alpha, infty}$, meaning $nu$ is either $alpha$ or $infty$.
Put ref{eq_expansion_for_R_G} back into ref{eq_R_G_before_expansion} which in turn goes into the expectation integral, we have
$$R_F(x) = sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} \
implies begin{aligned}[t]
E[X^r] &= r int_0^{infty} x^{r-1} sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} dd{x} \
&= r sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } int_0^{infty} x^{r-1} e^{-j,u(x)} dd{x}
end{aligned}$$
which is the desired expression.$quad Q.E.D.$
In case you're wondering, the exchange of the integral and summation is justified because the integrand-summand is positive and integrable. Please see e.g. this for more details or the relevant chapters in just any textbook on real analysis.
$endgroup$
answered Feb 2 at 12:44
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Thank you for all detail and answer. @ Lee David Chung Lin. You explained the things in very nice way.
$endgroup$
– J.H
Feb 4 at 10:29
add a comment |
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$begingroup$
Please load the page twice for the hyperlinks to work properly
$require{begingroup}begingrouprenewcommand{dd}[1]{,mathrm{d}#1}$Allow me to reiterate the setting:
$G$ is the baseline CDF for a non-negative random variable.
$R_G equiv 1 - G$ as the baseline survival function.- Define $u equiv -log R_G$, or equivalently $R_G = e^{-u}$
- Consider a random variable $X$ which CDF is $F = G^alpha$, where $alpha > 0 $ can be non-integer.
- When $alpha in mathbb{N}$ is an integer, then $X overset{d}{=}max { W_1,~W_2,~ldots~, W_{alpha}}$, namely, $X$ has the same distribution as the max of an iid set with $G$ being their common CDF. At any rate, $F$ is well-defined for any non-integer $alpha > 0$.
- Taking $G = 1 - R_G = 1 - e^{-u}$ yields $F = (1 - e^{-u})^{alpha}$, which upon differentiating yields the density $f$ as shown in the question post.
The $r$-th moment of $X$ by the simplest definition is
$$E[X^r] = int_0^{infty} x^r f(x) dd{x}$$
Meanwhile, there's a well-known relation between the expectation and the integral of survival function. In particular, see the second half of this answer and another anwser from the statistics site for a visualization. In the current notation, it is
$$ E[X^r] = r int_0^{infty} x^{r-1} R_F(x) dd{x} $$
where $R_F equiv 1 - F$ is the survival function of $X$ such that after invoking the given item #6 in the beginning, we have
$$R_F(x) = 1 - left(1 - e^{-u(x)}right)^{alpha} label{eq_R_G_before_expansion} tag*{Eq.(1)}$$
Now, apply the series expansion (generalized Binomial theorem) on this quantity $1 - e^{-u(x)}$, which magnitude is smaller than unity, raised to the $alpha$-th power.
begin{align}
left(1 - e^{-u(x)}right)^{alpha} &= sum_{k = 0}^{infty} { alpha choose k} left(- e^{-u(x)} right)^k \
&= sum_{k = 0}^{infty} (-1)^k frac{ alpha! }{k! (alpha - k)! } e^{-k,u(x)} \
&= 1 - alpha e^{-u(x)} + frac{ alpha (alpha - 1) }{ 2! } e^{-2u(x)} - frac{ alpha (alpha - 1)(alpha - 2) }{ 3! } e^{-3u(x)} + cdots \
&= 1 + sum_{j = 1}^{infty~~text{or}~~alpha} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} label{eq_expansion_for_R_G} tag*{Eq.(2)}
end{align}
The leading $1$ is pulled out in anticipation of the next step, and the shift in the lower summation limit automatically accommodates the expression $(alpha - j + 1)$.
There are two cases for the upper summation limit:
- when $alpha$ is non-integer, the upper-limit is $infty$ as the series goes on forever.
- When $alpha in mathbb{N}$, the series is just the ordinary binomial expansion that terminates at $j = alpha$ with a total of $alpha + 1$ terms.
One can succinctly write the summation as $displaystyle sum_{j = 1}^{nu}~$ , where $nu in {alpha, infty}$, meaning $nu$ is either $alpha$ or $infty$.
Put ref{eq_expansion_for_R_G} back into ref{eq_R_G_before_expansion} which in turn goes into the expectation integral, we have
$$R_F(x) = sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} \
implies begin{aligned}[t]
E[X^r] &= r int_0^{infty} x^{r-1} sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} dd{x} \
&= r sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } int_0^{infty} x^{r-1} e^{-j,u(x)} dd{x}
end{aligned}$$
which is the desired expression.$quad Q.E.D.$
In case you're wondering, the exchange of the integral and summation is justified because the integrand-summand is positive and integrable. Please see e.g. this for more details or the relevant chapters in just any textbook on real analysis.
$endgroup$
answered Feb 2 at 12:44
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Thank you for all detail and answer. @ Lee David Chung Lin. You explained the things in very nice way.
$endgroup$
– J.H
Feb 4 at 10:29
add a comment |
$begingroup$
Please load the page twice for the hyperlinks to work properly
$require{begingroup}begingrouprenewcommand{dd}[1]{,mathrm{d}#1}$Allow me to reiterate the setting:
$G$ is the baseline CDF for a non-negative random variable.
$R_G equiv 1 - G$ as the baseline survival function.- Define $u equiv -log R_G$, or equivalently $R_G = e^{-u}$
- Consider a random variable $X$ which CDF is $F = G^alpha$, where $alpha > 0 $ can be non-integer.
- When $alpha in mathbb{N}$ is an integer, then $X overset{d}{=}max { W_1,~W_2,~ldots~, W_{alpha}}$, namely, $X$ has the same distribution as the max of an iid set with $G$ being their common CDF. At any rate, $F$ is well-defined for any non-integer $alpha > 0$.
- Taking $G = 1 - R_G = 1 - e^{-u}$ yields $F = (1 - e^{-u})^{alpha}$, which upon differentiating yields the density $f$ as shown in the question post.
The $r$-th moment of $X$ by the simplest definition is
$$E[X^r] = int_0^{infty} x^r f(x) dd{x}$$
Meanwhile, there's a well-known relation between the expectation and the integral of survival function. In particular, see the second half of this answer and another anwser from the statistics site for a visualization. In the current notation, it is
$$ E[X^r] = r int_0^{infty} x^{r-1} R_F(x) dd{x} $$
where $R_F equiv 1 - F$ is the survival function of $X$ such that after invoking the given item #6 in the beginning, we have
$$R_F(x) = 1 - left(1 - e^{-u(x)}right)^{alpha} label{eq_R_G_before_expansion} tag*{Eq.(1)}$$
Now, apply the series expansion (generalized Binomial theorem) on this quantity $1 - e^{-u(x)}$, which magnitude is smaller than unity, raised to the $alpha$-th power.
begin{align}
left(1 - e^{-u(x)}right)^{alpha} &= sum_{k = 0}^{infty} { alpha choose k} left(- e^{-u(x)} right)^k \
&= sum_{k = 0}^{infty} (-1)^k frac{ alpha! }{k! (alpha - k)! } e^{-k,u(x)} \
&= 1 - alpha e^{-u(x)} + frac{ alpha (alpha - 1) }{ 2! } e^{-2u(x)} - frac{ alpha (alpha - 1)(alpha - 2) }{ 3! } e^{-3u(x)} + cdots \
&= 1 + sum_{j = 1}^{infty~~text{or}~~alpha} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} label{eq_expansion_for_R_G} tag*{Eq.(2)}
end{align}
The leading $1$ is pulled out in anticipation of the next step, and the shift in the lower summation limit automatically accommodates the expression $(alpha - j + 1)$.
There are two cases for the upper summation limit:
- when $alpha$ is non-integer, the upper-limit is $infty$ as the series goes on forever.
- When $alpha in mathbb{N}$, the series is just the ordinary binomial expansion that terminates at $j = alpha$ with a total of $alpha + 1$ terms.
One can succinctly write the summation as $displaystyle sum_{j = 1}^{nu}~$ , where $nu in {alpha, infty}$, meaning $nu$ is either $alpha$ or $infty$.
Put ref{eq_expansion_for_R_G} back into ref{eq_R_G_before_expansion} which in turn goes into the expectation integral, we have
$$R_F(x) = sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} \
implies begin{aligned}[t]
E[X^r] &= r int_0^{infty} x^{r-1} sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} dd{x} \
&= r sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } int_0^{infty} x^{r-1} e^{-j,u(x)} dd{x}
end{aligned}$$
which is the desired expression.$quad Q.E.D.$
In case you're wondering, the exchange of the integral and summation is justified because the integrand-summand is positive and integrable. Please see e.g. this for more details or the relevant chapters in just any textbook on real analysis.
$endgroup$
answered Feb 2 at 12:44
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Thank you for all detail and answer. @ Lee David Chung Lin. You explained the things in very nice way.
$endgroup$
– J.H
Feb 4 at 10:29
add a comment |
$begingroup$
Please load the page twice for the hyperlinks to work properly
$require{begingroup}begingrouprenewcommand{dd}[1]{,mathrm{d}#1}$Allow me to reiterate the setting:
$G$ is the baseline CDF for a non-negative random variable.
$R_G equiv 1 - G$ as the baseline survival function.- Define $u equiv -log R_G$, or equivalently $R_G = e^{-u}$
- Consider a random variable $X$ which CDF is $F = G^alpha$, where $alpha > 0 $ can be non-integer.
- When $alpha in mathbb{N}$ is an integer, then $X overset{d}{=}max { W_1,~W_2,~ldots~, W_{alpha}}$, namely, $X$ has the same distribution as the max of an iid set with $G$ being their common CDF. At any rate, $F$ is well-defined for any non-integer $alpha > 0$.
- Taking $G = 1 - R_G = 1 - e^{-u}$ yields $F = (1 - e^{-u})^{alpha}$, which upon differentiating yields the density $f$ as shown in the question post.
The $r$-th moment of $X$ by the simplest definition is
$$E[X^r] = int_0^{infty} x^r f(x) dd{x}$$
Meanwhile, there's a well-known relation between the expectation and the integral of survival function. In particular, see the second half of this answer and another anwser from the statistics site for a visualization. In the current notation, it is
$$ E[X^r] = r int_0^{infty} x^{r-1} R_F(x) dd{x} $$
where $R_F equiv 1 - F$ is the survival function of $X$ such that after invoking the given item #6 in the beginning, we have
$$R_F(x) = 1 - left(1 - e^{-u(x)}right)^{alpha} label{eq_R_G_before_expansion} tag*{Eq.(1)}$$
Now, apply the series expansion (generalized Binomial theorem) on this quantity $1 - e^{-u(x)}$, which magnitude is smaller than unity, raised to the $alpha$-th power.
begin{align}
left(1 - e^{-u(x)}right)^{alpha} &= sum_{k = 0}^{infty} { alpha choose k} left(- e^{-u(x)} right)^k \
&= sum_{k = 0}^{infty} (-1)^k frac{ alpha! }{k! (alpha - k)! } e^{-k,u(x)} \
&= 1 - alpha e^{-u(x)} + frac{ alpha (alpha - 1) }{ 2! } e^{-2u(x)} - frac{ alpha (alpha - 1)(alpha - 2) }{ 3! } e^{-3u(x)} + cdots \
&= 1 + sum_{j = 1}^{infty~~text{or}~~alpha} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} label{eq_expansion_for_R_G} tag*{Eq.(2)}
end{align}
The leading $1$ is pulled out in anticipation of the next step, and the shift in the lower summation limit automatically accommodates the expression $(alpha - j + 1)$.
There are two cases for the upper summation limit:
- when $alpha$ is non-integer, the upper-limit is $infty$ as the series goes on forever.
- When $alpha in mathbb{N}$, the series is just the ordinary binomial expansion that terminates at $j = alpha$ with a total of $alpha + 1$ terms.
One can succinctly write the summation as $displaystyle sum_{j = 1}^{nu}~$ , where $nu in {alpha, infty}$, meaning $nu$ is either $alpha$ or $infty$.
Put ref{eq_expansion_for_R_G} back into ref{eq_R_G_before_expansion} which in turn goes into the expectation integral, we have
$$R_F(x) = sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} \
implies begin{aligned}[t]
E[X^r] &= r int_0^{infty} x^{r-1} sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} dd{x} \
&= r sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } int_0^{infty} x^{r-1} e^{-j,u(x)} dd{x}
end{aligned}$$
which is the desired expression.$quad Q.E.D.$
In case you're wondering, the exchange of the integral and summation is justified because the integrand-summand is positive and integrable. Please see e.g. this for more details or the relevant chapters in just any textbook on real analysis.
$endgroup$
answered Feb 2 at 12:44
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
Please load the page twice for the hyperlinks to work properly
$require{begingroup}begingrouprenewcommand{dd}[1]{,mathrm{d}#1}$Allow me to reiterate the setting:
$G$ is the baseline CDF for a non-negative random variable.
$R_G equiv 1 - G$ as the baseline survival function.- Define $u equiv -log R_G$, or equivalently $R_G = e^{-u}$
- Consider a random variable $X$ which CDF is $F = G^alpha$, where $alpha > 0 $ can be non-integer.
- When $alpha in mathbb{N}$ is an integer, then $X overset{d}{=}max { W_1,~W_2,~ldots~, W_{alpha}}$, namely, $X$ has the same distribution as the max of an iid set with $G$ being their common CDF. At any rate, $F$ is well-defined for any non-integer $alpha > 0$.
- Taking $G = 1 - R_G = 1 - e^{-u}$ yields $F = (1 - e^{-u})^{alpha}$, which upon differentiating yields the density $f$ as shown in the question post.
The $r$-th moment of $X$ by the simplest definition is
$$E[X^r] = int_0^{infty} x^r f(x) dd{x}$$
Meanwhile, there's a well-known relation between the expectation and the integral of survival function. In particular, see the second half of this answer and another anwser from the statistics site for a visualization. In the current notation, it is
$$ E[X^r] = r int_0^{infty} x^{r-1} R_F(x) dd{x} $$
where $R_F equiv 1 - F$ is the survival function of $X$ such that after invoking the given item #6 in the beginning, we have
$$R_F(x) = 1 - left(1 - e^{-u(x)}right)^{alpha} label{eq_R_G_before_expansion} tag*{Eq.(1)}$$
Now, apply the series expansion (generalized Binomial theorem) on this quantity $1 - e^{-u(x)}$, which magnitude is smaller than unity, raised to the $alpha$-th power.
begin{align}
left(1 - e^{-u(x)}right)^{alpha} &= sum_{k = 0}^{infty} { alpha choose k} left(- e^{-u(x)} right)^k \
&= sum_{k = 0}^{infty} (-1)^k frac{ alpha! }{k! (alpha - k)! } e^{-k,u(x)} \
&= 1 - alpha e^{-u(x)} + frac{ alpha (alpha - 1) }{ 2! } e^{-2u(x)} - frac{ alpha (alpha - 1)(alpha - 2) }{ 3! } e^{-3u(x)} + cdots \
&= 1 + sum_{j = 1}^{infty~~text{or}~~alpha} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} label{eq_expansion_for_R_G} tag*{Eq.(2)}
end{align}
The leading $1$ is pulled out in anticipation of the next step, and the shift in the lower summation limit automatically accommodates the expression $(alpha - j + 1)$.
There are two cases for the upper summation limit:
- when $alpha$ is non-integer, the upper-limit is $infty$ as the series goes on forever.
- When $alpha in mathbb{N}$, the series is just the ordinary binomial expansion that terminates at $j = alpha$ with a total of $alpha + 1$ terms.
One can succinctly write the summation as $displaystyle sum_{j = 1}^{nu}~$ , where $nu in {alpha, infty}$, meaning $nu$ is either $alpha$ or $infty$.
Put ref{eq_expansion_for_R_G} back into ref{eq_R_G_before_expansion} which in turn goes into the expectation integral, we have
$$R_F(x) = sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} \
implies begin{aligned}[t]
E[X^r] &= r int_0^{infty} x^{r-1} sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } e^{-j,u(x)} dd{x} \
&= r sum_{j = 1}^{nu} (-1)^j frac{ alpha (alpha - 1) cdots (alpha - j + 1) }{ k! } int_0^{infty} x^{r-1} e^{-j,u(x)} dd{x}
end{aligned}$$
which is the desired expression.$quad Q.E.D.$
In case you're wondering, the exchange of the integral and summation is justified because the integrand-summand is positive and integrable. Please see e.g. this for more details or the relevant chapters in just any textbook on real analysis.
$endgroup$
answered Feb 2 at 12:44
Lee David Chung LinLee David Chung Lin
4,47841242
edited Feb 2 at 13:26
answered Feb 2 at 12:44
Lee David Chung LinLee David Chung Lin
4,47841242
answered Feb 2 at 12:44
Lee David Chung LinLee David Chung Lin
4,47841242
answered Feb 2 at 12:44
Lee David Chung LinLee David Chung Lin
4,47841242
4,47841242
$begingroup$
Thank you for all detail and answer. @ Lee David Chung Lin. You explained the things in very nice way.
$endgroup$
– J.H
Feb 4 at 10:29
add a comment |
$begingroup$
Thank you for all detail and answer. @ Lee David Chung Lin. You explained the things in very nice way.
$endgroup$
– J.H
Feb 4 at 10:29
$begingroup$
Thank you for all detail and answer. @ Lee David Chung Lin. You explained the things in very nice way.
$endgroup$
– J.H
Feb 4 at 10:29
$begingroup$
Thank you for all detail and answer. @ Lee David Chung Lin. You explained the things in very nice way.
$endgroup$
– J.H
Feb 4 at 10:29
add a comment |
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$begingroup$
@LeeDavidChungLin! yes [u(x)]'= du(x)/dx. It is derivative of u(x).
$endgroup$
– J.H
Jan 29 at 9:08
$begingroup$
$$ u(x)= -ln (R(x))$$ where R(x) is survival function. @LeeDavidChungLin
$endgroup$
– J.H
Jan 29 at 9:15
$begingroup$
Ok. Sorry for this. I do edit the question. @LeeDavidChungLin
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– J.H
Jan 29 at 11:24
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Yes I am still interested in answer. @LeeDavidChungLin
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– J.H
Feb 2 at 9:38
$begingroup$
@LeeDavidChungLin. our cdf is $$F(x) = (1-e^{-u(x)})^ {alpha} $$. I am also interested in expansion of its survival function (1-F(x)). which may be used to get moment.
$endgroup$
– J.H
Feb 2 at 9:52