Expectativa condicional da variável aleatória exponencial

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For a random variable XExp(λ) (E[X]=1λ) I feel intuitively that E[X|X>x] should equal x+E[X] since by the memoryless property the distribution of X|X>x is the same as that of X but shifted to the right by x.

However, I'm struggling to use the memoryless property to give a concrete proof. Any help is much appreciated.

Thanks.

mchen
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Hint: fX|X>a(x)=fX(xa) is the mathematical expression corresponding to "shifted to the right by a", and so
E[XX>a]=xfXX>a(x)dx=xfX(xa)dx.
Now do a change of variables on the integral on the right.
Dilip Sarwate
2
Note that X|X>x is a truncated distribution truncated below "x".Specially it is shifted exponential distribution and shifted exponential does not have memoryless property.
A.D

Respostas:

13

by the memoryless property the distribution of X|X>x is the same as that of X but shifted to the right by x.

Let fX(t) denote the probability density function (pdf) of X. Then, the mathematical formulation for what you correctly state namely, the conditional pdf of X given that {X>x} is the same as that of X but shifted to the right by x is that fXX>x(t)=fX(tx). Hence, E[XX>x], the expected value of X given that {X>x} is

E[XX>x]=tfXX>x(t)dt=tfX(tx)dt=(x+u)fX(u)duon substituting u=tx=x+E[X].
Note that we have not explicitly used the density of X in the calculation, and don't even need to integrate explicitly if we simply remember that (i) the area under a pdf is 1 and (ii) the definition of expected value of a continuous random variable in terms of its pdf.

Dilip Sarwate
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9

For x>0, the event {X>x} has probability P{X>x}=1FX(x)=eλx>0. Hence,

E[XX>x]=E[XI{X>x}]P{X>x},
but
E[XI{X>x}]=xtλeλtdt=()
(using Feynman's trick, vindicated by the Dominated Convergence Theorem, because it is fun)
()=λxddλ(eλt)dt=λddλxeλtdt
=λddλ(1λxλeλtdt)=λddλ(1λ(1FX(x)))
=λddλ(eλxλ)=(1λ+x)eλx,
which gives the desired result
E[XX>x]=1λ+x=E[X]+x.
Zen
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2
Although the use of Feynman's trick is interesting, why not just integrate by parts to get
xtλeλtdt=teλt|x+xeλtdt=(x+1λ)eλx?
Dilip Sarwate