Derivation of Markov's Inequality
$begingroup$
I have some problems understanding the derivation of Markov's inequality.
$$
begin{align}
EX &= int_{-infty}^{infty} xf_X(x)text{d}x \
&= int_{0}^infty xf_X(x)text{d}x & text{since } Xgt 0\
&geq int_{a}^infty xf_X(x)text{d}x &text{ for any }a gt 0\
&geqint_{a}^infty af_X(x)text{d}x & text{since } x > atext{ in the integrated region.}
end{align}
$$
I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference?
probability inequality
edited Dec 13 '18 at 19:06
Mutantoe
612513
asked Dec 13 '18 at 9:28
ChristianChristian
388
$endgroup$
add a comment |
$begingroup$
I have some problems understanding the derivation of Markov's inequality.
$$
begin{align}
EX &= int_{-infty}^{infty} xf_X(x)text{d}x \
&= int_{0}^infty xf_X(x)text{d}x & text{since } Xgt 0\
&geq int_{a}^infty xf_X(x)text{d}x &text{ for any }a gt 0\
&geqint_{a}^infty af_X(x)text{d}x & text{since } x > atext{ in the integrated region.}
end{align}
$$
I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference?
probability inequality
edited Dec 13 '18 at 19:06
Mutantoe
612513
asked Dec 13 '18 at 9:28
ChristianChristian
388
$endgroup$
add a comment |
$begingroup$
I have some problems understanding the derivation of Markov's inequality.
$$
begin{align}
EX &= int_{-infty}^{infty} xf_X(x)text{d}x \
&= int_{0}^infty xf_X(x)text{d}x & text{since } Xgt 0\
&geq int_{a}^infty xf_X(x)text{d}x &text{ for any }a gt 0\
&geqint_{a}^infty af_X(x)text{d}x & text{since } x > atext{ in the integrated region.}
end{align}
$$
I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference?
probability inequality
edited Dec 13 '18 at 19:06
Mutantoe
612513
asked Dec 13 '18 at 9:28
ChristianChristian
388
$endgroup$
I have some problems understanding the derivation of Markov's inequality.
$$
begin{align}
EX &= int_{-infty}^{infty} xf_X(x)text{d}x \
&= int_{0}^infty xf_X(x)text{d}x & text{since } Xgt 0\
&geq int_{a}^infty xf_X(x)text{d}x &text{ for any }a gt 0\
&geqint_{a}^infty af_X(x)text{d}x & text{since } x > atext{ in the integrated region.}
end{align}
$$
I don't understand the last step. Didn't we already use the fact that we were integrating over a larger region to write the inequality in the third line? Can someone help me with the difference?
probability inequality
probability inequality
edited Dec 13 '18 at 19:06
Mutantoe
612513
asked Dec 13 '18 at 9:28
ChristianChristian
388
edited Dec 13 '18 at 19:06
Mutantoe
612513
asked Dec 13 '18 at 9:28
ChristianChristian
388
edited Dec 13 '18 at 19:06
Mutantoe
612513
edited Dec 13 '18 at 19:06
Mutantoe
612513
edited Dec 13 '18 at 19:06
Mutantoe
612513
612513
asked Dec 13 '18 at 9:28
ChristianChristian
388
asked Dec 13 '18 at 9:28
ChristianChristian
388
asked Dec 13 '18 at 9:28
ChristianChristian
388
388
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
We could be more verbose in the third step as follows: Let $a > 0$. Then
$$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is where integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
$$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$
Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
edited Dec 13 '18 at 20:31
Acccumulation
7,0272619
answered Dec 13 '18 at 9:46
FalrachFalrach
1,664223
$endgroup$
add a comment |
$begingroup$
$ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.
answered Dec 13 '18 at 9:42
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
add a comment |
$begingroup$
The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.
answered Dec 13 '18 at 9:41
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
The last step uses the fact that the function
begin{align} g : [a,infty) rightarrow [0, infty)\
x mapsto x f_X(x)
end{align}
is always larger than
begin{align} h : [a,infty) rightarrow [0, infty)\
x mapsto a f_X(x)
end{align}
for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.
answered Dec 13 '18 at 9:40
JonathanJonathan
16312
$endgroup$
add a comment |
$begingroup$
No, there is a difference. Second to third step uses the fact that
$int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$
And
$int_0^a xf_X(x) dx>0$
Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration
Hope it is helpful
answered Dec 13 '18 at 9:47
MartundMartund
1,667213
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We could be more verbose in the third step as follows: Let $a > 0$. Then
$$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is where integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
$$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$
Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
edited Dec 13 '18 at 20:31
Acccumulation
7,0272619
answered Dec 13 '18 at 9:46
FalrachFalrach
1,664223
$endgroup$
add a comment |
$begingroup$
We could be more verbose in the third step as follows: Let $a > 0$. Then
$$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is where integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
$$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$
Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
edited Dec 13 '18 at 20:31
Acccumulation
7,0272619
answered Dec 13 '18 at 9:46
FalrachFalrach
1,664223
$endgroup$
add a comment |
$begingroup$
We could be more verbose in the third step as follows: Let $a > 0$. Then
$$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is where integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
$$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$
Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
edited Dec 13 '18 at 20:31
Acccumulation
7,0272619
answered Dec 13 '18 at 9:46
FalrachFalrach
1,664223
$endgroup$
We could be more verbose in the third step as follows: Let $a > 0$. Then
$$int_0^infty x f_X (x) text{d} x = int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x.$$
Now, since $f_X(x) geq 0$ and $xgeq 0 $ for $xin [0,a]$ we have that $int_0^a x f_X (x) text{d} x geq 0$ (This is where integrating over a larger/smaller region with a postitive integrand plays a role). Therefore
$$int_a^infty x f_X (x) text{d} x + int_0^a x f_X (x) text{d} x geq int_a^infty x f_X (x) text{d} x + 0 = int_a^infty x f_X (x) text{d} x$$
Now let us take a look on the last step: Here we want to use that for $xin[a,infty )$ we have $x geq a$ and therefore it holds that $x f_X (x) geq a f_X (x)$ (Here we just used that over our integral region we have a lower bound of the integrand). Altogether
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
edited Dec 13 '18 at 20:31
Acccumulation
7,0272619
answered Dec 13 '18 at 9:46
FalrachFalrach
1,664223
edited Dec 13 '18 at 20:31
Acccumulation
7,0272619
edited Dec 13 '18 at 20:31
Acccumulation
7,0272619
edited Dec 13 '18 at 20:31
Acccumulation
7,0272619
7,0272619
answered Dec 13 '18 at 9:46
FalrachFalrach
1,664223
answered Dec 13 '18 at 9:46
FalrachFalrach
1,664223
answered Dec 13 '18 at 9:46
FalrachFalrach
1,664223
1,664223
add a comment |
add a comment |
$begingroup$
$ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.
answered Dec 13 '18 at 9:42
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
add a comment |
$begingroup$
$ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.
answered Dec 13 '18 at 9:42
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
add a comment |
$begingroup$
$ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.
answered Dec 13 '18 at 9:42
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$ f geq g$ implies $int f geq int g$. Integrate $x f(x) I_{(a,infty)} geq a f(x) I_{(a,infty)}$ to get the last step.
answered Dec 13 '18 at 9:42
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 13 '18 at 9:42
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 13 '18 at 9:42
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Dec 13 '18 at 9:42
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
add a comment |
add a comment |
$begingroup$
The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.
answered Dec 13 '18 at 9:41
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.
answered Dec 13 '18 at 9:41
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.
answered Dec 13 '18 at 9:41
drhabdrhab
102k545136
$endgroup$
The last step applies: $$forall x;[0leq u(x)leq v(x)]impliesint u(x);dxleqint v(x);dx$$ This for $u(x)=af_X(x)mathbf1_{(a,infty)}(x)$ and $v(x)=xf_X(x)mathbf1_{(a,infty)}(x)$ where $ageq0$.
answered Dec 13 '18 at 9:41
drhabdrhab
102k545136
edited Dec 13 '18 at 9:48
answered Dec 13 '18 at 9:41
drhabdrhab
102k545136
answered Dec 13 '18 at 9:41
drhabdrhab
102k545136
answered Dec 13 '18 at 9:41
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
$begingroup$
The last step uses the fact that the function
begin{align} g : [a,infty) rightarrow [0, infty)\
x mapsto x f_X(x)
end{align}
is always larger than
begin{align} h : [a,infty) rightarrow [0, infty)\
x mapsto a f_X(x)
end{align}
for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.
answered Dec 13 '18 at 9:40
JonathanJonathan
16312
$endgroup$
add a comment |
$begingroup$
The last step uses the fact that the function
begin{align} g : [a,infty) rightarrow [0, infty)\
x mapsto x f_X(x)
end{align}
is always larger than
begin{align} h : [a,infty) rightarrow [0, infty)\
x mapsto a f_X(x)
end{align}
for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.
answered Dec 13 '18 at 9:40
JonathanJonathan
16312
$endgroup$
add a comment |
$begingroup$
The last step uses the fact that the function
begin{align} g : [a,infty) rightarrow [0, infty)\
x mapsto x f_X(x)
end{align}
is always larger than
begin{align} h : [a,infty) rightarrow [0, infty)\
x mapsto a f_X(x)
end{align}
for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.
answered Dec 13 '18 at 9:40
JonathanJonathan
16312
$endgroup$
The last step uses the fact that the function
begin{align} g : [a,infty) rightarrow [0, infty)\
x mapsto x f_X(x)
end{align}
is always larger than
begin{align} h : [a,infty) rightarrow [0, infty)\
x mapsto a f_X(x)
end{align}
for (a > 0) in its domain. As you correctly state, the proof uses the fact that you integrate a non-negative function over a smaller region in the second-to-last inequality, not in the last inequality.
answered Dec 13 '18 at 9:40
JonathanJonathan
16312
answered Dec 13 '18 at 9:40
JonathanJonathan
16312
answered Dec 13 '18 at 9:40
JonathanJonathan
16312
answered Dec 13 '18 at 9:40
JonathanJonathan
16312
16312
add a comment |
add a comment |
$begingroup$
No, there is a difference. Second to third step uses the fact that
$int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$
And
$int_0^a xf_X(x) dx>0$
Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration
Hope it is helpful
answered Dec 13 '18 at 9:47
MartundMartund
1,667213
$endgroup$
add a comment |
$begingroup$
No, there is a difference. Second to third step uses the fact that
$int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$
And
$int_0^a xf_X(x) dx>0$
Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration
Hope it is helpful
answered Dec 13 '18 at 9:47
MartundMartund
1,667213
$endgroup$
add a comment |
$begingroup$
No, there is a difference. Second to third step uses the fact that
$int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$
And
$int_0^a xf_X(x) dx>0$
Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration
Hope it is helpful
answered Dec 13 '18 at 9:47
MartundMartund
1,667213
$endgroup$
No, there is a difference. Second to third step uses the fact that
$int_0^infty xf_X(x)dx = int_0^a xf_X(x)dx+int_a^infty xf_X(x)dx$
And
$int_0^a xf_X(x) dx>0$
Last step uses that $int_a^infty xf_X(x)dx>int_a^infty af_X(x)dx$, because $x>a$ in these limits of integration
Hope it is helpful
answered Dec 13 '18 at 9:47
MartundMartund
1,667213
answered Dec 13 '18 at 9:47
MartundMartund
1,667213
answered Dec 13 '18 at 9:47
MartundMartund
1,667213
answered Dec 13 '18 at 9:47
MartundMartund
1,667213
1,667213
add a comment |
add a comment |
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