How to prove that if $E(X^2) < infty$ then $E(X) < infty$?
$begingroup$
How to prove that if $E(X^2) < infty$ then $E(X) < infty$?
Here is my attempt:
It's easy to show that if $E(X^2) < infty$ then $E(X) < infty$ when $X^2 ge X$ (by using the monotonicity of expectation).
But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as
$$
Eg(x) = int |g(x)|f(x) < infty?
$$
probability-theory expected-value
asked Dec 23 '18 at 18:31
DianneDianne
62
$endgroup$
add a comment |
$begingroup$
How to prove that if $E(X^2) < infty$ then $E(X) < infty$?
Here is my attempt:
It's easy to show that if $E(X^2) < infty$ then $E(X) < infty$ when $X^2 ge X$ (by using the monotonicity of expectation).
But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as
$$
Eg(x) = int |g(x)|f(x) < infty?
$$
probability-theory expected-value
asked Dec 23 '18 at 18:31
DianneDianne
62
$endgroup$
2
$begingroup$
$X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
$endgroup$
– Yanko
Dec 23 '18 at 18:34
4
$begingroup$
$$|X|leqslant X^2+1quad text{(QED)}$$
$endgroup$
– Did
Dec 23 '18 at 19:01
add a comment |
$begingroup$
How to prove that if $E(X^2) < infty$ then $E(X) < infty$?
Here is my attempt:
It's easy to show that if $E(X^2) < infty$ then $E(X) < infty$ when $X^2 ge X$ (by using the monotonicity of expectation).
But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as
$$
Eg(x) = int |g(x)|f(x) < infty?
$$
probability-theory expected-value
asked Dec 23 '18 at 18:31
DianneDianne
62
$endgroup$
How to prove that if $E(X^2) < infty$ then $E(X) < infty$?
Here is my attempt:
It's easy to show that if $E(X^2) < infty$ then $E(X) < infty$ when $X^2 ge X$ (by using the monotonicity of expectation).
But how do I prove the case when $X^2 < X$, using only the definition of finiteness of expectation as
$$
Eg(x) = int |g(x)|f(x) < infty?
$$
probability-theory expected-value
probability-theory expected-value
asked Dec 23 '18 at 18:31
DianneDianne
62
asked Dec 23 '18 at 18:31
DianneDianne
62
edited Dec 27 '18 at 5:35
user587192
asked Dec 23 '18 at 18:31
DianneDianne
62
asked Dec 23 '18 at 18:31
DianneDianne
62
asked Dec 23 '18 at 18:31
DianneDianne
62
62
2
$begingroup$
$X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
$endgroup$
– Yanko
Dec 23 '18 at 18:34
4
$begingroup$
$$|X|leqslant X^2+1quad text{(QED)}$$
$endgroup$
– Did
Dec 23 '18 at 19:01
add a comment |
2
$begingroup$
$X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
$endgroup$
– Yanko
Dec 23 '18 at 18:34
4
$begingroup$
$$|X|leqslant X^2+1quad text{(QED)}$$
$endgroup$
– Did
Dec 23 '18 at 19:01
2
2
$begingroup$
$X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
$endgroup$
– Yanko
Dec 23 '18 at 18:34
$begingroup$
$X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
$endgroup$
– Yanko
Dec 23 '18 at 18:34
4
4
$begingroup$
$$|X|leqslant X^2+1quad text{(QED)}$$
$endgroup$
– Did
Dec 23 '18 at 19:01
$begingroup$
$$|X|leqslant X^2+1quad text{(QED)}$$
$endgroup$
– Did
Dec 23 '18 at 19:01
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The one reached onto you by the comment of Did is my favourite.
Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$
answered Dec 23 '18 at 20:15
drhabdrhab
102k545136
$endgroup$
$begingroup$
(+1) A nice quantization of the difference using a common formula for the variance.
$endgroup$
– robjohn♦
Dec 23 '18 at 23:06
1
$begingroup$
@robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
$endgroup$
– drhab
Dec 24 '18 at 6:00
$begingroup$
Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
$endgroup$
– robjohn♦
Dec 24 '18 at 9:17
add a comment |
$begingroup$
By Jensen inequity for $g(x) = x^ 2$,
$$
g(mathbb{E}(X)) le E( g(X) ),
$$
hence,
$$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
$$
thus
$$
mathbb{E}(X) < infty
$$
answered Dec 23 '18 at 18:42
V. VancakV. Vancak
11.2k2926
$endgroup$
1
$begingroup$
(+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
$endgroup$
– robjohn♦
Dec 23 '18 at 19:04
add a comment |
$begingroup$
Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
$$
E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
$$
answered Dec 23 '18 at 18:49
robjohn♦robjohn
268k27309635
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The one reached onto you by the comment of Did is my favourite.
Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$
answered Dec 23 '18 at 20:15
drhabdrhab
102k545136
$endgroup$
$begingroup$
(+1) A nice quantization of the difference using a common formula for the variance.
$endgroup$
– robjohn♦
Dec 23 '18 at 23:06
1
$begingroup$
@robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
$endgroup$
– drhab
Dec 24 '18 at 6:00
$begingroup$
Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
$endgroup$
– robjohn♦
Dec 24 '18 at 9:17
add a comment |
$begingroup$
The one reached onto you by the comment of Did is my favourite.
Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$
answered Dec 23 '18 at 20:15
drhabdrhab
102k545136
$endgroup$
$begingroup$
(+1) A nice quantization of the difference using a common formula for the variance.
$endgroup$
– robjohn♦
Dec 23 '18 at 23:06
1
$begingroup$
@robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
$endgroup$
– drhab
Dec 24 '18 at 6:00
$begingroup$
Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
$endgroup$
– robjohn♦
Dec 24 '18 at 9:17
add a comment |
$begingroup$
The one reached onto you by the comment of Did is my favourite.
Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$
answered Dec 23 '18 at 20:15
drhabdrhab
102k545136
$endgroup$
The one reached onto you by the comment of Did is my favourite.
Another one is: $$mathbb EX^2-(mathbb EX)^2=mathbb E(X-mathbb EX)^2geq0$$implying that: $$(mathbb EX)^2leqmathbb EX^2<infty$$
answered Dec 23 '18 at 20:15
drhabdrhab
102k545136
edited Dec 23 '18 at 20:29
answered Dec 23 '18 at 20:15
drhabdrhab
102k545136
answered Dec 23 '18 at 20:15
drhabdrhab
102k545136
answered Dec 23 '18 at 20:15
drhabdrhab
102k545136
102k545136
$begingroup$
(+1) A nice quantization of the difference using a common formula for the variance.
$endgroup$
– robjohn♦
Dec 23 '18 at 23:06
1
$begingroup$
@robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
$endgroup$
– drhab
Dec 24 '18 at 6:00
$begingroup$
Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
$endgroup$
– robjohn♦
Dec 24 '18 at 9:17
add a comment |
$begingroup$
(+1) A nice quantization of the difference using a common formula for the variance.
$endgroup$
– robjohn♦
Dec 23 '18 at 23:06
1
$begingroup$
@robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
$endgroup$
– drhab
Dec 24 '18 at 6:00
$begingroup$
Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
$endgroup$
– robjohn♦
Dec 24 '18 at 9:17
$begingroup$
(+1) A nice quantization of the difference using a common formula for the variance.
$endgroup$
– robjohn♦
Dec 23 '18 at 23:06
$begingroup$
(+1) A nice quantization of the difference using a common formula for the variance.
$endgroup$
– robjohn♦
Dec 23 '18 at 23:06
1
1
$begingroup$
@robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
$endgroup$
– drhab
Dec 24 '18 at 6:00
$begingroup$
@robjohn Thank you! It has a minus though. It carries some circular reference. Before we can use what is symbolized in $mathbb EX$ (so that we can use the symbol) we should first have proved that that is a legal thing to do. This legalization however is proving that $mathbb E|X|<infty$. In this context on base of $mathbb EX^2<infty$. This can be done by e.g. the hint of Did.
$endgroup$
– drhab
Dec 24 '18 at 6:00
$begingroup$
Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
$endgroup$
– robjohn♦
Dec 24 '18 at 9:17
$begingroup$
Yes, indeed. $mathbb{E}(X)$ appears in two of the terms, which makes it hard to isolate. Maybe the OP won't notice ;-)
$endgroup$
– robjohn♦
Dec 24 '18 at 9:17
add a comment |
$begingroup$
By Jensen inequity for $g(x) = x^ 2$,
$$
g(mathbb{E}(X)) le E( g(X) ),
$$
hence,
$$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
$$
thus
$$
mathbb{E}(X) < infty
$$
answered Dec 23 '18 at 18:42
V. VancakV. Vancak
11.2k2926
$endgroup$
1
$begingroup$
(+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
$endgroup$
– robjohn♦
Dec 23 '18 at 19:04
add a comment |
$begingroup$
By Jensen inequity for $g(x) = x^ 2$,
$$
g(mathbb{E}(X)) le E( g(X) ),
$$
hence,
$$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
$$
thus
$$
mathbb{E}(X) < infty
$$
answered Dec 23 '18 at 18:42
V. VancakV. Vancak
11.2k2926
$endgroup$
1
$begingroup$
(+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
$endgroup$
– robjohn♦
Dec 23 '18 at 19:04
add a comment |
$begingroup$
By Jensen inequity for $g(x) = x^ 2$,
$$
g(mathbb{E}(X)) le E( g(X) ),
$$
hence,
$$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
$$
thus
$$
mathbb{E}(X) < infty
$$
answered Dec 23 '18 at 18:42
V. VancakV. Vancak
11.2k2926
$endgroup$
By Jensen inequity for $g(x) = x^ 2$,
$$
g(mathbb{E}(X)) le E( g(X) ),
$$
hence,
$$(mathbb{E}X )^2 le mathbb{E}X^2 < infty,
$$
thus
$$
mathbb{E}(X) < infty
$$
answered Dec 23 '18 at 18:42
V. VancakV. Vancak
11.2k2926
answered Dec 23 '18 at 18:42
V. VancakV. Vancak
11.2k2926
answered Dec 23 '18 at 18:42
V. VancakV. Vancak
11.2k2926
answered Dec 23 '18 at 18:42
V. VancakV. Vancak
11.2k2926
11.2k2926
1
$begingroup$
(+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
$endgroup$
– robjohn♦
Dec 23 '18 at 19:04
add a comment |
1
$begingroup$
(+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
$endgroup$
– robjohn♦
Dec 23 '18 at 19:04
1
1
$begingroup$
(+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
$endgroup$
– robjohn♦
Dec 23 '18 at 19:04
$begingroup$
(+1) Usually, I am the one using Jensen when others have used Cauchy-Schwarz. It's nice to switch things up once in a while.
$endgroup$
– robjohn♦
Dec 23 '18 at 19:04
add a comment |
$begingroup$
Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
$$
E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
$$
answered Dec 23 '18 at 18:49
robjohn♦robjohn
268k27309635
$endgroup$
add a comment |
$begingroup$
Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
$$
E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
$$
answered Dec 23 '18 at 18:49
robjohn♦robjohn
268k27309635
$endgroup$
add a comment |
$begingroup$
Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
$$
E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
$$
answered Dec 23 '18 at 18:49
robjohn♦robjohn
268k27309635
$endgroup$
Cauchy-Schwarz (or Cauchy-Bunyakovsky-Schwarz) says
$$
E(X)^2le E!left(X^2right)E(1)=E!left(X^2right)
$$
answered Dec 23 '18 at 18:49
robjohn♦robjohn
268k27309635
edited Dec 23 '18 at 18:58
answered Dec 23 '18 at 18:49
robjohn♦robjohn
268k27309635
answered Dec 23 '18 at 18:49
robjohn♦robjohn
268k27309635
answered Dec 23 '18 at 18:49
robjohn♦robjohn
268k27309635
268k27309635
add a comment |
add a comment |
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2
$begingroup$
$X^2leq X$ if and only if $Xleq 1$. Moreover $int 1 dx = 1$ so you're good.
$endgroup$
– Yanko
Dec 23 '18 at 18:34
4
$begingroup$
$$|X|leqslant X^2+1quad text{(QED)}$$
$endgroup$
– Did
Dec 23 '18 at 19:01