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Hi Noel,
Please help me with this one :
Suppose E( X )< infinity, mu=E( X ) Fix c<d .
Let ~X= c , x, d accordingly as [X<c],[c<=X<=d],[X>d] ,set ~mu=E(~X)
Prove: E( ~X  ~mu^r )<=E(X mu^r) for all r>= 1
Regards,
Shuva
PS: I solved the problem for mu=~mu but didn't need the condition r>=1, ie true for any r
Not very sure whether the problem is correct .If wrong can you get an counter example.
You many have to use Cr Inequality , Minkowski's , or Jensen's or Whatever .
Tony Wong <tw813@...> wrote:
hello all,
I have the following question which I hope someone can
help with:
Given X_t is a brownian motion with drift rate u and
volatility sigma, and X_0 = a >0, define T(b) with b>a
to be the first time X_t hits the level b. Now define
Y_t = X_{min(t, T(b)}. (i.e. Y_t=X_t if T(b) is
greater than t and Y_t=b for all t>= T(b) ). For each
fixed t, Y_t has a mixed distribution with a point
mass at b and a continuous density on (infinity, b).
My question is: how can one derive the density part??
(For the case where X_t is the standard brownian
motion, this question is not too hard.) Also, can
someone tell me from which book, I could get an answer
for this question??
Many thanks,
Tony
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Hi Noel,
Can you please help me with this
Suppose EX< infinity, mu=E(X). Fix c<d ,
Let ~X = c,X,d according as [X<c], [c<=X<=d], [X>d]
Set ~mu=E(~X)
Show that E ( ~X  ~mu^r ) <=E ( X  mu^r ) for all r>=1
Regards,
Shuva
PS: I have been able to do it for mu = ~mu ( did not need the condn r>=1here )but couldn't for the case mu not equals ~mu.I am noy sure whether the problem is correct so can you please provide me with a counter example if you think it is not true.
You many need to use Cr Inequality ,Jensen or Whatever .....

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Here's a heuristic sketch of a very special case. Let's suppose that
c=\infty, that X has a "nice" density ("nice" here means whatever
it needs to mean so that all my dubious differentiations are
justified), and that r=2k for some positive integer k.
Then ~X=min(X,d). Define g(t)=E[min(X,t)] and observe that
g'(t) = E[1_{X > t}] = P(X > t).
Also note that ~mu=g(d).
Now define h(t) = E[(min(X,t)  g(t))^{2k}], so that E~X  ~mu^r =
h(d). Note that
h'(t) = 2k E[(min(X,t)  g(t))^{2k1} (1_{X > t}  g'(t))].
If we let Y be an independent copy of X, then
1_{X > t}  g'(t)
1_{X > t} P(Y < t)  1_{X < t} P(Y > t)
= E[1_{X > t > Y}X]  E[1_{X < t < Y}X].
Hence,
h'(t) = 2k E[(min(X,t)  g(t))^{2k1} 1_{X > t > Y}]
 2k E[(min(X,t)  g(t))^{2k1} 1_{X < t < Y}]
= 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
 2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
Since 2k1 is odd,
(X  g(t))^{2k1} 1_{X < t < Y}
<= (t  g(t))^{2k1} 1_{X < t < Y}.
Therefore,
h'(t) >= 2k(t  g(t))^{2k1} E[1_{X > t > Y}  1_{X < t < Y}].
But since X and Y are iid, this expectation is 0, so h is an
increasing function in t. Thus,
E~X  ~mu^r = h(d)
<= lim_{t>\infty} h(t)
= EX  mu^r. QED
This argument is so convoluted, I don't have much confidence in it.
(Where's my mistake?) Even if it is correct, hopefully someone else
(Noel?) can provide something more transparent.
 In probability@yahoogroups.com, shuva gupta <shuvagupta@y...>
wrote:>
condn r>=1here )but couldn't for the case mu not equals ~mu.I am
>
> Hi Noel,
>
> Can you please help me with this
>
> Suppose EX< infinity, mu=E(X). Fix c<d ,
>
> Let ~X = c,X,d according as [X<c], [c<=X<=d], [X>d]
>
> Set ~mu=E(~X)
>
> Show that E ( ~X  ~mu^r ) <=E ( X  mu^r ) for all r>=1
>
> Regards,
>
> Shuva
>
> PS: I have been able to do it for mu = ~mu ( did not need the
noy sure whether the problem is correct so can you please provide me
with a counter example if you think it is not true.>
> You many need to use Cr Inequality ,Jensen or Whatever .....
>
>
>
>
>
>
>
>
>
>
>
>
> 
> Do you Yahoo!?
> Check out the new Yahoo! Front Page. www.yahoo.com/a
>
> [Nontext portions of this message have been removed] 0 Attachment
Well, here's one mistake:
I wrote
> Since 2k1 is odd,
What I wanted to say was
>
> (X  g(t))^{2k1} 1_{X < t < Y}
> <= (t  g(t))^{2k1} 1_{X < t < Y}.
> Since 2k1 is odd,
But that's clearly false. I then used this false "fact" to show that
>
> (X  g(t))^{2k1} 1_{X < t < Y}
> <= (t  g(t))^{2k1} 1_{X > t > Y}.
h'(t)>=0, which finished the "proof".
But I think it's salvagable. Here's another way to show h'(t)>=0.
Start with what we know:
h'(t) = 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
 2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
Since X and Y are iid, I can interchange them in the first
expectation, giving
h'(t) = 2k E[(t  g(t))^{2k1} 1_{X < t < Y}]
 2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
Now we use the fact that (X  g(t))^{2k1} <= (t  g(t))^{2k1}
whenever X < t to conclude that h'(t)>=0.
Any other glaring mistakes? (Besides the lack of rigor?) :)
 In probability@yahoogroups.com, "jason1990" <jason1990@y...>
wrote:>
that
>
> Here's a heuristic sketch of a very special case. Let's suppose
> c=\infty, that X has a "nice" density ("nice" here means whatever
=
> it needs to mean so that all my dubious differentiations are
> justified), and that r=2k for some positive integer k.
>
> Then ~X=min(X,d). Define g(t)=E[min(X,t)] and observe that
>
> g'(t) = E[1_{X > t}] = P(X > t).
>
> Also note that ~mu=g(d).
>
> Now define h(t) = E[(min(X,t)  g(t))^{2k}], so that E~X  ~mu^r
> h(d). Note that
it.
>
> h'(t) = 2k E[(min(X,t)  g(t))^{2k1} (1_{X > t}  g'(t))].
>
> If we let Y be an independent copy of X, then
>
> 1_{X > t}  g'(t)
> 1_{X > t} P(Y < t)  1_{X < t} P(Y > t)
> = E[1_{X > t > Y}X]  E[1_{X < t < Y}X].
>
> Hence,
>
> h'(t) = 2k E[(min(X,t)  g(t))^{2k1} 1_{X > t > Y}]
>  2k E[(min(X,t)  g(t))^{2k1} 1_{X < t < Y}]
> = 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
>  2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
>
> Since 2k1 is odd,
>
> (X  g(t))^{2k1} 1_{X < t < Y}
> <= (t  g(t))^{2k1} 1_{X < t < Y}.
>
> Therefore,
>
> h'(t) >= 2k(t  g(t))^{2k1} E[1_{X > t > Y}  1_{X < t < Y}].
>
> But since X and Y are iid, this expectation is 0, so h is an
> increasing function in t. Thus,
>
> E~X  ~mu^r = h(d)
> <= lim_{t>\infty} h(t)
> = EX  mu^r. QED
>
> This argument is so convoluted, I don't have much confidence in
> (Where's my mistake?) Even if it is correct, hopefully someone
else
> (Noel?) can provide something more transparent.
>
>  In probability@yahoogroups.com, shuva gupta <shuvagupta@y...>
> wrote:
> >
> >
> > Hi Noel,
> >
> > Can you please help me with this
> >
> > Suppose EX< infinity, mu=E(X). Fix c<d ,
> >
> > Let ~X = c,X,d according as [X<c], [c<=X<=d], [X>d]
> >
> > Set ~mu=E(~X)
> >
> > Show that E ( ~X  ~mu^r ) <=E ( X  mu^r ) for all r>=1 0 Attachment
> This argument is so convoluted, I don't have much confidence in it.
Jason,
> (Where's my mistake?) Even if it is correct, hopefully someone else
> (Noel?) can provide something more transparent.
I find this argument very impressive (there is a lot of ideas in it).
I am a bit uneasy about differentiation under E[...] but I am
confident that this is not a flaw in your proof (i.e. I am confident
every equality could be rigorously justified). I know you have
mentioned a mistake (cf your next post), but I haven't seen any on
first reading :) So I am gonna go through your next post now.
Noel. 0 Attachment
> > Since 2k1 is odd,
Since (2k1) is odd, x>x^(2k1)is nondecreasing on R
> >
> > (X  g(t))^{2k1} 1_{X < t < Y}
> > <= (t  g(t))^{2k1} 1_{X < t < Y}.
and since X  g(t) <= t  g(t) on {X < t}, the inequality
you wrote seems fine to me.
I also think this inequality allows you to conclude
that h'(t) >= 0.
h'(t)
= 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
 2k E[(X  g(t))^{2k1} 1_{X < t < Y}]>= 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
 2k E[(t  g(t))^{2k1} 1_{X < t < Y}]
= 2k(tg(t))^{2k1}E[1_{X > t > Y}  1_{X < t < Y}]
=0
which is pretty much what you wrote in your first post.
What am I missing?
> What I wanted to say was
I wonder why you wanted to write this in the first place.
>
> > Since 2k1 is odd,
> >
> > (X  g(t))^{2k1} 1_{X < t < Y}
> > <= (t  g(t))^{2k1} 1_{X > t > Y}.
>
> But that's clearly false.
> But I think it's salvagable. Here's another way to show
This seems to me like the same as what you wrote in the first place,
>h'(t)>=0. Start with what we know:
>
> h'(t) = 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
>  2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
>
> Since X and Y are iid, I can interchange them in the first
> expectation, giving
>
> h'(t) = 2k E[(t  g(t))^{2k1} 1_{X < t < Y}]
>  2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
>
> Now we use the fact that (X  g(t))^{2k1} <= (t  g(t))^{2k1}
> whenever X < t to conclude that h'(t)>=0.
>
> Any other glaring mistakes? (Besides the lack of rigor?) :)
except you use iid property, before taking inequalities.
Any way, it seems to me your proof is good. I certainly haven't
seen a flaw.
Noel. 0 Attachment
Yes, you're right. I thought about this in the morning on the bus
and again this evening before going to a movie. I had two lines of
reasoning in my head at once and when I reread my post, I confused
myself.
About differentiating under the expectation, something occurred to
me at the theater. For any nonnegative random variable Y and any
r>0, E[Y^r]=\int_0^\infty{rz^{r1}P(Y>z)dz}. So suppose we're given
a real t and we want to compute E[min(X,t)]. Let Y=tmin(X,t). Then
Y is nonnegative and we can apply the above to get
E[min(X,t)] = t  \int_{\infty}^t{P(X < z)dz}.
We can differentiate this with no problem. Something similar should
be possible with the other expectation. I don't think this is
necessary to justify the differentiation, but it makes me wonder
whether the whole differentiation approach is unnecessary.
Another thing that is curious: the only property of the map x>x^
{2k} that was used is the fact that it has a nondecreasing
derivative. In other words, it is convex. So perhaps the original
poster's claim is true not only for functions of the form x>x^r
where r>=1, but for all convex functions. If so, then maybe Jensen's
inequality would be useful in creating a simpler proof.
I just feel that it shouldn't be this hard.
 In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
wrote:>
place,
>
> > > Since 2k1 is odd,
> > >
> > > (X  g(t))^{2k1} 1_{X < t < Y}
> > > <= (t  g(t))^{2k1} 1_{X < t < Y}.
>
> Since (2k1) is odd, x>x^(2k1)is nondecreasing on R
> and since X  g(t) <= t  g(t) on {X < t}, the inequality
> you wrote seems fine to me.
>
> I also think this inequality allows you to conclude
> that h'(t) >= 0.
>
> h'(t)
> = 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
>  2k E[(X  g(t))^{2k1} 1_{X < t < Y}]
> >= 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
>  2k E[(t  g(t))^{2k1} 1_{X < t < Y}]
> = 2k(tg(t))^{2k1}E[1_{X > t > Y}  1_{X < t < Y}]
> =0
>
> which is pretty much what you wrote in your first post.
> What am I missing?
>
> > What I wanted to say was
> >
> > > Since 2k1 is odd,
> > >
> > > (X  g(t))^{2k1} 1_{X < t < Y}
> > > <= (t  g(t))^{2k1} 1_{X > t > Y}.
> >
> > But that's clearly false.
>
> I wonder why you wanted to write this in the first place.
>
>
>
> > But I think it's salvagable. Here's another way to show
> >h'(t)>=0. Start with what we know:
> >
> > h'(t) = 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
> >  2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
> >
> > Since X and Y are iid, I can interchange them in the first
> > expectation, giving
> >
> > h'(t) = 2k E[(t  g(t))^{2k1} 1_{X < t < Y}]
> >  2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
> >
> > Now we use the fact that (X  g(t))^{2k1} <= (t  g(t))^{2k1}
> > whenever X < t to conclude that h'(t)>=0.
> >
> > Any other glaring mistakes? (Besides the lack of rigor?) :)
>
> This seems to me like the same as what you wrote in the first
> except you use iid property, before taking inequalities.
>
> Any way, it seems to me your proof is good. I certainly haven't
> seen a flaw.
>
> Noel. 0 Attachment
> About differentiating under the expectation, something occurred to
Very good. I agree.
> me at the theater. For any nonnegative random variable Y and any
> r>0, E[Y^r]=\int_0^\infty{rz^{r1}P(Y>z)dz}. So suppose we're given
> a real t and we want to compute E[min(X,t)]. Let Y=tmin(X,t). Then
> Y is nonnegative and we can apply the above to get
>
> E[min(X,t)] = t  \int_{\infty}^t{P(X < z)dz}.
>
> We can differentiate this with no problem.
Yes.
> Something similar should be possible with the other
Yes.
> expectation.
> Another thing that is curious: the only property of the map x>x^
Well, I'd be happy already to crack this for r in [0,+oo[
> {2k} that was used is the fact that it has a nondecreasing
> derivative. In other words, it is convex. So perhaps the original
> poster's claim is true not only for functions of the form x>x^r
> where r>=1, but for all convex functions.
> If so, then maybe Jensen's inequality would be useful in
Yes, I am guessing there should be a simpler proof, but I
>creating a simpler proof.
can't find it. I have been looking for a while now.
Even looked for counterexample for r=2.
I am going to give up soon :)
Noel. 0 Attachment
I am very sorry Shuva,
I have been stuck for 2 hours on this.
I need to move on, otherwise I ll go crazy.
I think Jason has a good chance to find a complete proof.
Noel. 0 Attachment
Thanks Jason Noel and Myriam,
Actually the problem has a very simple solution if we exploit the fact that
g(x)=x^r, r>=1 is a convex function.
ie
we use the property g(x)g(y)>=(xy)*g'(y) when g(.)is convex.
Put x=XE(X)
y= ~X  E(~X) Then take expectation on both sides .
Regards,
Shuva
PS A friend of mine found this solution in the book Probability by Chow and Teicher 1st Edition ( pg 102103)
jason1990 <jason1990@...> wrote:
Yes, you're right. I thought about this in the morning on the bus
and again this evening before going to a movie. I had two lines of
reasoning in my head at once and when I reread my post, I confused
myself.
About differentiating under the expectation, something occurred to
me at the theater. For any nonnegative random variable Y and any
r>0, E[Y^r]=\int_0^\infty{rz^{r1}P(Y>z)dz}. So suppose we're given
a real t and we want to compute E[min(X,t)]. Let Y=tmin(X,t). Then
Y is nonnegative and we can apply the above to get
E[min(X,t)] = t  \int_{\infty}^t{P(X < z)dz}.
We can differentiate this with no problem. Something similar should
be possible with the other expectation. I don't think this is
necessary to justify the differentiation, but it makes me wonder
whether the whole differentiation approach is unnecessary.
Another thing that is curious: the only property of the map x>x^
{2k} that was used is the fact that it has a nondecreasing
derivative. In other words, it is convex. So perhaps the original
poster's claim is true not only for functions of the form x>x^r
where r>=1, but for all convex functions. If so, then maybe Jensen's
inequality would be useful in creating a simpler proof.
I just feel that it shouldn't be this hard.
 In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
wrote:>
place,
>
> > > Since 2k1 is odd,
> > >
> > > (X  g(t))^{2k1} 1_{X < t < Y}
> > > <= (t  g(t))^{2k1} 1_{X < t < Y}.
>
> Since (2k1) is odd, x>x^(2k1)is nondecreasing on R
> and since X  g(t) <= t  g(t) on {X < t}, the inequality
> you wrote seems fine to me.
>
> I also think this inequality allows you to conclude
> that h'(t) >= 0.
>
> h'(t)
> = 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
>  2k E[(X  g(t))^{2k1} 1_{X < t < Y}]
> >= 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
>  2k E[(t  g(t))^{2k1} 1_{X < t < Y}]
> = 2k(tg(t))^{2k1}E[1_{X > t > Y}  1_{X < t < Y}]
> =0
>
> which is pretty much what you wrote in your first post.
> What am I missing?
>
> > What I wanted to say was
> >
> > > Since 2k1 is odd,
> > >
> > > (X  g(t))^{2k1} 1_{X < t < Y}
> > > <= (t  g(t))^{2k1} 1_{X > t > Y}.
> >
> > But that's clearly false.
>
> I wonder why you wanted to write this in the first place.
>
>
>
> > But I think it's salvagable. Here's another way to show
> >h'(t)>=0. Start with what we know:
> >
> > h'(t) = 2k E[(t  g(t))^{2k1} 1_{X > t > Y}]
> >  2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
> >
> > Since X and Y are iid, I can interchange them in the first
> > expectation, giving
> >
> > h'(t) = 2k E[(t  g(t))^{2k1} 1_{X < t < Y}]
> >  2k E[(X  g(t))^{2k1} 1_{X < t < Y}].
> >
> > Now we use the fact that (X  g(t))^{2k1} <= (t  g(t))^{2k1}
> > whenever X < t to conclude that h'(t)>=0.
> >
> > Any other glaring mistakes? (Besides the lack of rigor?) :)
>
> This seems to me like the same as what you wrote in the first
> except you use iid property, before taking inequalities.
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>
> Any way, it seems to me your proof is good. I certainly haven't
> seen a flaw.
>
> Noel.

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Is it obvious that E[(xy)g'(y)]>=0 for this choice of x and y?
I am not sure I understand.
Noel.
> g(x)=x^r, r>=1 is a convex function.
> ie
> we use the property g(x)g(y)>=(xy)*g'(y) when g(.)is convex.
> Put x=XE(X)
> y= ~X  E(~X) Then take expectation on both sides . 0 Attachment
For what it's worth, it's not obvious at all to me. But maybe we're
both missing something. The book is apparently "Probability theory :
independence, interchangeability, martingales" by Yuan Shih Chow,
Henry Teicher. I have a request on that book just to see what's
going on. If someone can enlighten me while I wait, that would be
great.
 In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
wrote:>
>
>
> Is it obvious that E[(xy)g'(y)]>=0 for this choice of x and y?
> I am not sure I understand.
>
> Noel.
>
>
> > g(x)=x^r, r>=1 is a convex function.
> > ie
> > we use the property g(x)g(y)>=(xy)*g'(y) when g(.)is convex.
> > Put x=XE(X)
> > y= ~X  E(~X) Then take expectation on both sides . 0 Attachment
Sorry folks, for the confusion,
Sorry Jason you are right the book is
"Probability theory : independence, interchangeability, martingales" by Yuan Shih Chow,
Henry Teicher.
If G(.) is a convex fxn then this propert hold :
G(x)G(y) >=(xy)G'r(y) { G'r(y):= rt hand derivative of G(y) at the pt y moreover G'(.) is an increasing function.}
Now take x=XE(X) y=~XE(~X)
therefore
G(XE(X))G(~XE(~X))>=(XE(X)(~XE(~X))G'r(~XE(~X))
Now it can be shown that ((XE(X)(~XE(~X))G'r(~XE(~X)) is >=
(XE(X)(~XE(~X))*K ( where K is some constant, the detailed argument is in the book , using the fact that G'r(.) is increasing and also the fact the function a(x)=x~x is monotonically incraesing (where ~x=a if x<=a, =x if a=<x <=b and =b if x>=b; a<b)
Thus we have
G(XE(X))G(~XE(~X))>=(XE(X)(~XE(~X))*K
Now taking expectation on both sides we have
E(G(XE(X))) E(G(~XE(~X)))>=E((XE(X)(~XE(~X))*K)
Since E((XE(X)(~XE(~X))*K)=0 thus we have
E(G(XE(X))) E(G(~XE(~X)))>=0
Take G(x)=x^r (r>=1) and we get the desired result ie
E(XE(X)^r)>=E(~XE(~X)^r)
If it is still not clear please let me know and I will be glad to put in more details ( or maybe I can also scan a couple of relevant pages from the book and send it across only if you ask ....I dont want to unnecessary overload the inboxes:))))))))))
Best Regards,
Shuva
jason1990 <jason1990@...> wrote:
For what it's worth, it's not obvious at all to me. But maybe we're
both missing something. The book is apparently "Probability theory :
independence, interchangeability, martingales" by Yuan Shih Chow,
Henry Teicher. I have a request on that book just to see what's
going on. If someone can enlighten me while I wait, that would be
great.
 In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
wrote:>
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>
>
> Is it obvious that E[(xy)g'(y)]>=0 for this choice of x and y?
> I am not sure I understand.
>
> Noel.
>
>
> > g(x)=x^r, r>=1 is a convex function.
> > ie
> > we use the property g(x)g(y)>=(xy)*g'(y) when g(.)is convex.
> > Put x=XE(X)
> > y= ~X  E(~X) Then take expectation on both sides .
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[Nontext portions of this message have been removed] 0 Attachment
> Now it can be shown that:
((XE(X)(~XE(~X))G'r(~XE(~X)) >= (XE(X)(~XE(~X))*K
> where K is some constant, [...] using the fact that G'r(.) is
I feel very dumb here. Tried but failed. Will someone take me
>increasing and [...] a(x)=x~x is monotonically increasing
out of my misery? I know (~XE(~X)) is bounded, and since G'
is nondecreasing, G'r(~XE(~X)) is also bounded. But somehow
I can't manage to conclude (and use the hint about a(x))
I have been spending so much time on this. I may as well
go to the end, so it won't have been for nothing :)
Noel. 0 Attachment
I think I see it now. Define
f(t) = t  ~t  E(X) + E(~X) and
g(t) = G'r(~t  E(~X)).
Both functions are nondecreasing, and we can find t_0 such that
f(t) <= 0 for t <= t_0 and
f(t) >= 0 for t >= t_0.
Hence, if X >= t_0, then since f(X) >= 0 and g(X) >= g(t_0), we have
f(X)g(X) >= f(X)g(t_0).
Also, if X <= t_0, then f(X) <= 0 and g(X) <= g(t_0), so
f(X)g(X) >= f(X)g(t_0).
So I guess we take K = g(t_0). I think this works. I wouldn't call
it obvious, though.
 In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
wrote:>
>
> > Now it can be shown that:
>
> ((XE(X)(~XE(~X))G'r(~XE(~X)) >= (XE(X)(~XE(~X))*K
>
> > where K is some constant, [...] using the fact that G'r(.) is
> >increasing and [...] a(x)=x~x is monotonically increasing
>
> I feel very dumb here. Tried but failed. Will someone take me
> out of my misery? I know (~XE(~X)) is bounded, and since G'
> is nondecreasing, G'r(~XE(~X)) is also bounded. But somehow
> I can't manage to conclude (and use the hint about a(x))
>
> I have been spending so much time on this. I may as well
> go to the end, so it won't have been for nothing :)
>
> Noel. 0 Attachment
Thank you very much Jason. This looks very good to me,
and is a huge relief :)
Noel.
> I think I see it now. Define
>
> f(t) = t  ~t  E(X) + E(~X) and
> g(t) = G'r(~t  E(~X)).
>
> Both functions are nondecreasing, and we can find t_0 such that
>
> f(t) <= 0 for t <= t_0 and
> f(t) >= 0 for t >= t_0.
>
> Hence, if X >= t_0, then since f(X) >= 0 and g(X) >= g(t_0), we have
>
> f(X)g(X) >= f(X)g(t_0).
>
> Also, if X <= t_0, then f(X) <= 0 and g(X) <= g(t_0), so
>
> f(X)g(X) >= f(X)g(t_0).
>
> So I guess we take K = g(t_0).
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