relation between moments of discrete random variable and CDF
$begingroup$
Consider a discrete positive random variable, say X. This link nicely shows that
begin{equation}
E[X] = sum_{k=0}^{infty} (1-F(k))
end{equation}
Moreover,
begin{equation}
E[X^2] = sum_{k=0}^{infty} (2k+1)(1-F(k))
end{equation}
I am puzzled on how to obtain the factor "(2k+1)" in above expression. It would be nice, when there were a relation between F(X) and X for higher order moments.
probability-distributions
asked Jan 16 at 17:53
Alexander K.Alexander K.
1
$endgroup$
add a comment |
$begingroup$
Consider a discrete positive random variable, say X. This link nicely shows that
begin{equation}
E[X] = sum_{k=0}^{infty} (1-F(k))
end{equation}
Moreover,
begin{equation}
E[X^2] = sum_{k=0}^{infty} (2k+1)(1-F(k))
end{equation}
I am puzzled on how to obtain the factor "(2k+1)" in above expression. It would be nice, when there were a relation between F(X) and X for higher order moments.
probability-distributions
asked Jan 16 at 17:53
Alexander K.Alexander K.
1
$endgroup$
$begingroup$
Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
$endgroup$
– angryavian
Jan 16 at 18:10
$begingroup$
Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
$endgroup$
– Alexander K.
Jan 16 at 20:00
add a comment |
$begingroup$
Consider a discrete positive random variable, say X. This link nicely shows that
begin{equation}
E[X] = sum_{k=0}^{infty} (1-F(k))
end{equation}
Moreover,
begin{equation}
E[X^2] = sum_{k=0}^{infty} (2k+1)(1-F(k))
end{equation}
I am puzzled on how to obtain the factor "(2k+1)" in above expression. It would be nice, when there were a relation between F(X) and X for higher order moments.
probability-distributions
asked Jan 16 at 17:53
Alexander K.Alexander K.
1
$endgroup$
Consider a discrete positive random variable, say X. This link nicely shows that
begin{equation}
E[X] = sum_{k=0}^{infty} (1-F(k))
end{equation}
Moreover,
begin{equation}
E[X^2] = sum_{k=0}^{infty} (2k+1)(1-F(k))
end{equation}
I am puzzled on how to obtain the factor "(2k+1)" in above expression. It would be nice, when there were a relation between F(X) and X for higher order moments.
probability-distributions
probability-distributions
asked Jan 16 at 17:53
Alexander K.Alexander K.
1
asked Jan 16 at 17:53
Alexander K.Alexander K.
1
asked Jan 16 at 17:53
Alexander K.Alexander K.
1
asked Jan 16 at 17:53
Alexander K.Alexander K.
1
asked Jan 16 at 17:53
Alexander K.Alexander K.
1
1
$begingroup$
Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
$endgroup$
– angryavian
Jan 16 at 18:10
$begingroup$
Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
$endgroup$
– Alexander K.
Jan 16 at 20:00
add a comment |
$begingroup$
Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
$endgroup$
– angryavian
Jan 16 at 18:10
$begingroup$
Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
$endgroup$
– Alexander K.
Jan 16 at 20:00
$begingroup$
Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
$endgroup$
– angryavian
Jan 16 at 18:10
$begingroup$
Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
$endgroup$
– angryavian
Jan 16 at 18:10
$begingroup$
Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
$endgroup$
– Alexander K.
Jan 16 at 20:00
$begingroup$
Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
$endgroup$
– Alexander K.
Jan 16 at 20:00
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer in your link gives you the explanation already...
If $G$ is the CDF of $X^2$ then the first formula yields
$$E[X^2] = sum_{k = 0}^infty (1 - G(k))
= (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.
answered Jan 16 at 20:40
angryavianangryavian
42.5k23481
$endgroup$
add a comment |
$begingroup$
thank you for the hint. From this post, it is now clear that one has to rewrite
begin{equation}
E[X^2] = sum_k k^2 P(X=k)
end{equation}
as
begin{equation}
E[X^2] = sum_k H(k) P(X>k)
end{equation}
where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
Setting
begin{equation}
sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
end{equation}
one can easily get the term H(k).
answered Jan 18 at 16:48
Alexander K.Alexander K.
1
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer in your link gives you the explanation already...
If $G$ is the CDF of $X^2$ then the first formula yields
$$E[X^2] = sum_{k = 0}^infty (1 - G(k))
= (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.
answered Jan 16 at 20:40
angryavianangryavian
42.5k23481
$endgroup$
add a comment |
$begingroup$
The answer in your link gives you the explanation already...
If $G$ is the CDF of $X^2$ then the first formula yields
$$E[X^2] = sum_{k = 0}^infty (1 - G(k))
= (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.
answered Jan 16 at 20:40
angryavianangryavian
42.5k23481
$endgroup$
add a comment |
$begingroup$
The answer in your link gives you the explanation already...
If $G$ is the CDF of $X^2$ then the first formula yields
$$E[X^2] = sum_{k = 0}^infty (1 - G(k))
= (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.
answered Jan 16 at 20:40
angryavianangryavian
42.5k23481
$endgroup$
The answer in your link gives you the explanation already...
If $G$ is the CDF of $X^2$ then the first formula yields
$$E[X^2] = sum_{k = 0}^infty (1 - G(k))
= (1-G(0)) + (1-G(1)) + (1-G(2)) + cdots.$$
Since $X$ takes integer values, we have $X^2 in {0^2, 1^2, 2^2, 3^3, ldots}$, so $G(1)=G(2)=G(3)=F(1)$ and $G(4) = G(5) = G(6) = G(7) = G(8) = F(2)$ and so on.
Writing all the $G$ terms (in the above sum) in terms of $F$ and collecting like terms yields the $(2k+1)(1-F(k))$ expression.
answered Jan 16 at 20:40
angryavianangryavian
42.5k23481
answered Jan 16 at 20:40
angryavianangryavian
42.5k23481
answered Jan 16 at 20:40
angryavianangryavian
42.5k23481
answered Jan 16 at 20:40
angryavianangryavian
42.5k23481
42.5k23481
add a comment |
add a comment |
$begingroup$
thank you for the hint. From this post, it is now clear that one has to rewrite
begin{equation}
E[X^2] = sum_k k^2 P(X=k)
end{equation}
as
begin{equation}
E[X^2] = sum_k H(k) P(X>k)
end{equation}
where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
Setting
begin{equation}
sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
end{equation}
one can easily get the term H(k).
answered Jan 18 at 16:48
Alexander K.Alexander K.
1
$endgroup$
add a comment |
$begingroup$
thank you for the hint. From this post, it is now clear that one has to rewrite
begin{equation}
E[X^2] = sum_k k^2 P(X=k)
end{equation}
as
begin{equation}
E[X^2] = sum_k H(k) P(X>k)
end{equation}
where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
Setting
begin{equation}
sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
end{equation}
one can easily get the term H(k).
answered Jan 18 at 16:48
Alexander K.Alexander K.
1
$endgroup$
add a comment |
$begingroup$
thank you for the hint. From this post, it is now clear that one has to rewrite
begin{equation}
E[X^2] = sum_k k^2 P(X=k)
end{equation}
as
begin{equation}
E[X^2] = sum_k H(k) P(X>k)
end{equation}
where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
Setting
begin{equation}
sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
end{equation}
one can easily get the term H(k).
answered Jan 18 at 16:48
Alexander K.Alexander K.
1
$endgroup$
thank you for the hint. From this post, it is now clear that one has to rewrite
begin{equation}
E[X^2] = sum_k k^2 P(X=k)
end{equation}
as
begin{equation}
E[X^2] = sum_k H(k) P(X>k)
end{equation}
where H(k) is the unknown term. When we write the elements of the sum as matrix, then the sum in the first equation corresponds to column-wise summation, while that of the second equation corresponds to row-wise summation.
Setting
begin{equation}
sum_k H(k) P(X>k) = sum_k k^2 P(X=k)
end{equation}
one can easily get the term H(k).
answered Jan 18 at 16:48
Alexander K.Alexander K.
1
answered Jan 18 at 16:48
Alexander K.Alexander K.
1
answered Jan 18 at 16:48
Alexander K.Alexander K.
1
answered Jan 18 at 16:48
Alexander K.Alexander K.
1
1
add a comment |
add a comment |
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$begingroup$
Doesn't the answer to the question in your link explain where the $2k+1$ comes from?
$endgroup$
– angryavian
Jan 16 at 18:10
$begingroup$
Please, what link are you referring to? The only link, namely stats.stackexchange.com/a/90591/25936 , points to an application using above expression.
$endgroup$
– Alexander K.
Jan 16 at 20:00