Re: E[X]E[exp(-at)] >= E[X*exp(-at)]?
- From: hrubin@xxxxxxxxxxxxxxxxxxxx (Herman Rubin)
- Date: 4 Oct 2007 21:47:54 -0400
In article <1191528351.961000.93220@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Yecloud <yecloud@xxxxxxxxx> wrote:
On Oct 4, 3:46 pm, hru...@xxxxxxxxxxxxxxxxxxxx (Herman Rubin) wrote:
In article <1191517449.363752.244...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Yecloud <yecl...@xxxxxxxxx> wrote:
On Oct 4, 12:27 pm, s...@xxxxxxxxxxxxxx wrote:
On 4 Oct, 16:50, Yecloud <yecl...@xxxxxxxxx> wrote:
Hi, all,
Suppose X>=0, t>=0 are random variables with arbitrary distributions
and a > 0, do we have
E[X]E[exp(-at)] >= E[X*exp(-at)]?
The random variable Y = exp(-at) can be any random variable
in the open interval (0,1). There is no reason it should
be positively correlated with X.
"negatively" correlated?
In fact, t=X gives a counterexample if X is non-trivial.
The above example is not a counterexample. Another example, if t=X
and X~exp(lambda), nontrivial, the inequality holds. Could you
construct a counterexample?
Ok, I had the signs mixed up. Take X = 1 - exp(-at),
or any decreasing function of t.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@xxxxxxxxxxxxxxx Phone: (765)494-6054 FAX: (765)494-0558
.
- References:
- E[X]E[exp(-at)] >= E[X*exp(-at)]?
- From: Yecloud
- Re: E[X]E[exp(-at)] >= E[X*exp(-at)]?
- From: Yecloud
- Re: E[X]E[exp(-at)] >= E[X*exp(-at)]?
- From: Herman Rubin
- Re: E[X]E[exp(-at)] >= E[X*exp(-at)]?
- From: Yecloud
- E[X]E[exp(-at)] >= E[X*exp(-at)]?
- Prev by Date: Re: Function on an infinite number of real variables
- Next by Date: Re: Bourbaki's street address?
- Previous by thread: Re: E[X]E[exp(-at)] >= E[X*exp(-at)]?
- Next by thread: Linear Independence
- Index(es):
Relevant Pages
|
