Expectativa condicional da variável aleatória exponencial

13

For a random variable $X\sim \text{Exp}(\lambda)$ ( $\mathbb{E}[X] = \frac{1}{\lambda}$ ) I feel intuitively that $\mathbb{E}[X|X > x]$ should equal $x + \mathbb{E}[X]$ 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.

random-variable exponential conditional-expectation

— mchen
fonte

Hint:

f_{X | X > a} (x) = f_{X} (x - a)

$f_{X|X > a}(x) = f_X(x-a)$ is the mathematical expression corresponding to "shifted to the right by

a

$a$ ", and so

E [X ∣ X > a] = \int_{- \infty}^{\infty} x f_{X ∣ X > a} (x) d x = \int_{- \infty}^{\infty} x f_{X} (x - a) d x .

$E[X\mid X>a] = \int_{-\infty}^\infty xf_{X\mid X>a}(x)\,\mathrm dx= \int_{-\infty}^\infty xf_X(x-a)\,\mathrm dx.$ Now do a change of variables on the integral on the right.

— Dilip Sarwate

2

Note that

X | X > x

$X|X>x$ is a truncated distribution truncated below "

x

$x$ ".Specially it is shifted exponential distribution and shifted exponential does not have memoryless property.

— A.D

13

$\ldots$ 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 $f_X(t)$ 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 $f_{X \mid X > x}(t) = f_X(t-x)$ . Hence, $E[X\mid X > x]$ , the expected value of $X$ given that $\{X > x\}$ is

\begin{aligned} E [X ∣ X > x] & = \int_{- \infty}^{\infty} t f_{X ∣ X > x} (t) d t \\ = \int_{- \infty}^{\infty} t f_{X} (t - x) d t \\ = \int_{- \infty}^{\infty} (x + u) f_{X} (u) d u & on substituting u = t - x \\ = x + E [X] . \end{aligned}

$\begin{align} E[X\mid X > x] &= \int_{-\infty}^\infty t f_{X \mid X > x}(t)\,\mathrm dt\\ &= \int_{-\infty}^\infty t f_X(t-x)\,\mathrm dt\\ &= \int_{-\infty}^\infty (x+u) f_X(u)\,\mathrm du &\scriptstyle{\text{on substituting}~u = t-x}\\ &= x + E[X]. \end{align}$ Note that we have not explicitly used the density of

X

$X$ in the calculation, and don't even need to integrate explicitly if we simply remember that (i) the area under a pdf is

1

$1$ and (ii) the definition of expected value of a continuous random variable in terms of its pdf.

— Dilip Sarwate
fonte

9

For $x>0$ , the event $\{X>x\}$ has probability $P\{X>x\}=1-F_X(x)=e^{-\lambda x} > 0$ . Hence,

E [X ∣ X > x] = \frac{E [X I_{{X > x}}]}{P {X > x}},

$\newcommand{\E}{\mathbb{E}} \E[X\mid X> x] = \frac{\E[X\,I_{\{X>x\}}]}{P\{X>x\}} \, ,$ but

E [X I_{{X > x}}] = \int_{x}^{\infty} t λ e^{- λ t} d t = (*)

$\E[X\,I_{\{X>x\}}] = \int_x^\infty t\,\lambda\,e^{-\lambda t}\,dt = (*)$ (using Feynman's trick, vindicated by the Dominated Convergence Theorem, because it is fun)

(*) = - λ \int_{x}^{\infty} \frac{d}{d λ} (e^{- λ t}) d t = - λ \frac{d}{d λ} \int_{x}^{\infty} e^{- λ t} d t

$(*) = -\lambda \int_x^\infty \frac{d}{d\lambda}(e^{-\lambda t}) \,dt = -\lambda \frac{d}{d\lambda} \int_x^\infty e^{-\lambda t} \,dt$

= - λ \frac{d}{d λ} (\frac{1}{λ} \int_{x}^{\infty} λ e^{- λ t} d t) = - λ \frac{d}{d λ} (\frac{1}{λ} (1 - F_{X} (x)))

$= -\lambda \frac{d}{d\lambda} \left(\frac{1}{\lambda} \int_x^\infty \lambda\,e^{-\lambda t} \,dt\right) = -\lambda\frac{d}{d\lambda}\left(\frac{1}{\lambda}\left(1 - F_X(x)\right)\right)$

= - λ \frac{d}{d λ} (\frac{e^{- λ x}}{λ}) = (\frac{1}{λ} + x) e^{- λ x},

$= -\lambda\frac{d}{d\lambda}\left(\frac{e^{-\lambda x}}{\lambda}\right) = \left(\frac{1}{\lambda}+x\right)e^{-\lambda x} \, ,$ which gives the desired result

E [X ∣ X > x] = \frac{1}{λ} + x = E [X] + x .

$\E[X\mid X>x] = \frac{1}{\lambda}+x = \E[X] + x \, .$

— Zen
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2

Although the use of Feynman's trick is interesting, why not just integrate by parts to get

\int_{x}^{\infty} t λ e^{- λ t} d t = - t e^{- λ t} |_{x}^{\infty} + \int_{x}^{\infty} e^{- λ t} d t = (x + \frac{1}{λ}) e^{- λ x} ?

$\int_x^\infty t\lambda e^{-\lambda t}\,dt = -t e^{-\lambda t}\biggr |_x^\infty + \int_x^\infty e^{-\lambda t}\,dt = \left(x + \frac{1}{\lambda}\right)e^{-\lambda x}?$

— Dilip Sarwate