E[X]E[exp(-at)] >= E[X*exp(-at)]?

From: Yecloud <yecloud@xxxxxxxxx>
Date: Thu, 04 Oct 2007 08:50:09 -0700

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)]?

Thanks,
Cloud

.

Follow-Ups:
- Re: E[X]E[exp(-at)] >= E[X*exp(-at)]?
  - From: se16

Prev by Date: Re: Two results of set geometry
Next by Date: Re: Two results of set geometry
Previous by thread: Reference needed
Next by thread: Re: E[X]E[exp(-at)] >= E[X*exp(-at)]?
Index(es):
- Date
- Thread