Derivation of Markov's Inequality
up vote
2
down vote
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I have some problems understanding the derivation of Markov's inequality.
$EX = int_{-infty}^{infty} xf_X(x)text{d}x$
$= int_{0}^infty xf_X(x)text{d}x$ since $X$ is positive.
$geq int_{a}^infty xf_X(x)text{d}x$ for any positive $a$.
$geq int_{a}^infty af_X(x)text{d}x$ since $ x > a$ in the integrated region.
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 2 hours ago
Falrach
1,545223
asked 2 hours ago
Christian
347
add a comment |
up vote
2
down vote
favorite
I have some problems understanding the derivation of Markov's inequality.
$EX = int_{-infty}^{infty} xf_X(x)text{d}x$
$= int_{0}^infty xf_X(x)text{d}x$ since $X$ is positive.
$geq int_{a}^infty xf_X(x)text{d}x$ for any positive $a$.
$geq int_{a}^infty af_X(x)text{d}x$ since $ x > a$ in the integrated region.
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 2 hours ago
Falrach
1,545223
asked 2 hours ago
Christian
347
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have some problems understanding the derivation of Markov's inequality.
$EX = int_{-infty}^{infty} xf_X(x)text{d}x$
$= int_{0}^infty xf_X(x)text{d}x$ since $X$ is positive.
$geq int_{a}^infty xf_X(x)text{d}x$ for any positive $a$.
$geq int_{a}^infty af_X(x)text{d}x$ since $ x > a$ in the integrated region.
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 2 hours ago
Falrach
1,545223
asked 2 hours ago
Christian
347
I have some problems understanding the derivation of Markov's inequality.
$EX = int_{-infty}^{infty} xf_X(x)text{d}x$
$= int_{0}^infty xf_X(x)text{d}x$ since $X$ is positive.
$geq int_{a}^infty xf_X(x)text{d}x$ for any positive $a$.
$geq int_{a}^infty af_X(x)text{d}x$ since $ x > a$ in the integrated region.
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 2 hours ago
Falrach
1,545223
asked 2 hours ago
Christian
347
edited 2 hours ago
Falrach
1,545223
asked 2 hours ago
Christian
347
edited 2 hours ago
Falrach
1,545223
edited 2 hours ago
Falrach
1,545223
edited 2 hours ago
Falrach
1,545223
1,545223
asked 2 hours ago
Christian
347
asked 2 hours ago
Christian
347
asked 2 hours ago
Christian
347
347
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
up vote
3
down vote
accepted
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 were 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
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
answered 2 hours ago
Falrach
1,545223
More verbose $ne$ More precise.
– Did
2 hours ago
Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
– Falrach
2 hours ago
You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
– Did
1 hour ago
1
Thank you Falrach. Your longer explanation was very helpful for me.
– Christian
56 mins ago
add a comment |
up vote
3
down vote
$ 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 2 hours ago
Kavi Rama Murthy
46.5k31854
add a comment |
up vote
3
down vote
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 2 hours ago
drhab
95.7k543126
add a comment |
up vote
2
down vote
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 2 hours ago
Jonathan
16012
add a comment |
up vote
2
down vote
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 2 hours ago
Martund
1,122211
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
up vote
3
down vote
accepted
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 were 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
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
answered 2 hours ago
Falrach
1,545223
More verbose $ne$ More precise.
– Did
2 hours ago
Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
– Falrach
2 hours ago
You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
– Did
1 hour ago
1
Thank you Falrach. Your longer explanation was very helpful for me.
– Christian
56 mins ago
add a comment |
up vote
3
down vote
accepted
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 were 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
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
answered 2 hours ago
Falrach
1,545223
More verbose $ne$ More precise.
– Did
2 hours ago
Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
– Falrach
2 hours ago
You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
– Did
1 hour ago
1
Thank you Falrach. Your longer explanation was very helpful for me.
– Christian
56 mins ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
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 were 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
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
answered 2 hours ago
Falrach
1,545223
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 were 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
$$int_a^infty x f_X (x) text{d} x geq int_a^infty a f_X (x) text{d} x = a int_a^infty f_X (x) text{d} x$$
answered 2 hours ago
Falrach
1,545223
edited 2 hours ago
answered 2 hours ago
Falrach
1,545223
answered 2 hours ago
Falrach
1,545223
answered 2 hours ago
Falrach
1,545223
1,545223
More verbose $ne$ More precise.
– Did
2 hours ago
Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
– Falrach
2 hours ago
You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
– Did
1 hour ago
1
Thank you Falrach. Your longer explanation was very helpful for me.
– Christian
56 mins ago
add a comment |
More verbose $ne$ More precise.
– Did
2 hours ago
Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
– Falrach
2 hours ago
You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
– Did
1 hour ago
1
Thank you Falrach. Your longer explanation was very helpful for me.
– Christian
56 mins ago
More verbose $ne$ More precise.
– Did
2 hours ago
More verbose $ne$ More precise.
– Did
2 hours ago
Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
– Falrach
2 hours ago
Hey Did I would be really thankful for constructive criticism. As you can see there a several short answers. If they are more helpful than mine it is good as it is.
– Falrach
2 hours ago
You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
– Did
1 hour ago
You are the one who saw fit to start your answer with the claim that it would be more precise; it is not.
– Did
1 hour ago
1
1
Thank you Falrach. Your longer explanation was very helpful for me.
– Christian
56 mins ago
Thank you Falrach. Your longer explanation was very helpful for me.
– Christian
56 mins ago
add a comment |
up vote
3
down vote
$ 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 2 hours ago
Kavi Rama Murthy
46.5k31854
add a comment |
up vote
3
down vote
$ 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 2 hours ago
Kavi Rama Murthy
46.5k31854
add a comment |
up vote
3
down vote
up vote
3
down vote
$ 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 2 hours ago
Kavi Rama Murthy
46.5k31854
$ 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 2 hours ago
Kavi Rama Murthy
46.5k31854
answered 2 hours ago
Kavi Rama Murthy
46.5k31854
answered 2 hours ago
Kavi Rama Murthy
46.5k31854
answered 2 hours ago
Kavi Rama Murthy
46.5k31854
46.5k31854
add a comment |
add a comment |
up vote
3
down vote
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 2 hours ago
drhab
95.7k543126
add a comment |
up vote
3
down vote
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 2 hours ago
drhab
95.7k543126
add a comment |
up vote
3
down vote
up vote
3
down vote
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 2 hours ago
drhab
95.7k543126
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 2 hours ago
drhab
95.7k543126
edited 2 hours ago
answered 2 hours ago
drhab
95.7k543126
answered 2 hours ago
drhab
95.7k543126
answered 2 hours ago
drhab
95.7k543126
95.7k543126
add a comment |
add a comment |
up vote
2
down vote
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 2 hours ago
Jonathan
16012
add a comment |
up vote
2
down vote
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 2 hours ago
Jonathan
16012
add a comment |
up vote
2
down vote
up vote
2
down vote
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 2 hours ago
Jonathan
16012
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 2 hours ago
Jonathan
16012
answered 2 hours ago
Jonathan
16012
answered 2 hours ago
Jonathan
16012
answered 2 hours ago
Jonathan
16012
16012
add a comment |
add a comment |
up vote
2
down vote
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 2 hours ago
Martund
1,122211
add a comment |
up vote
2
down vote
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 2 hours ago
Martund
1,122211
add a comment |
up vote
2
down vote
up vote
2
down vote
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 2 hours ago
Martund
1,122211
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 2 hours ago
Martund
1,122211
answered 2 hours ago
Martund
1,122211
answered 2 hours ago
Martund
1,122211
answered 2 hours ago
Martund
1,122211
1,122211
add a comment |
add a comment |
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