EEL6507sp09L29

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EEL6507 Spring 2009, Lecture 29, Wednesday 2009/03/25 (Notes created by Yonggang Liu)

  • Problem of Residual life

Problem and Paradox

One H event happens in the infinite process
One H event happens in the infinite process
  • What is the pdf for Y?

Assume exponential distribution of X, paradox appears (Guess one):

  1. H event shows up half way through on average, so since <X>= \frac {1}{\lambda}, we get <Y>= \frac {1}{2\lambda}.
  2. Since A is memoryless, <Y>= \frac {1}{\lambda}.

Preferred Selection

  • Big things / values have proportionally higher probability to be selected.

We can prove this in common sense: when a satellite falls, it has much higher probability to fall into the ocean than to fall into the lakes or rivers, because the ocean is much larger than the lakes and rivers.

Mathematically, suppose there are 4 samples, A1,A2,A3,A4:

A1,A2,A3,A4 are within intervals X1,X2,X3,X4
A1,A2,A3,A4 are within intervals X1,X2,X3,X4
\begin{alignat}{2} Prob[choosing X_1] &=\frac {X_1} {X_1+X_2+X_3+X_4}\\ Prob[choosing X_2] &=\frac {X_2} {X_1+X_2+X_3+X_4}\\ Prob[choosing X_3] &=\frac {X_3} {X_1+X_2+X_3+X_4}\\ Prob[choosing X_4] &=\frac {X_4} {X_1+X_2+X_3+X_4} \end{alignat}

So, the probability of choosing Xi is proportional to the value of Xi.

Distribution of Choosing block X

pdf of choosing block with size x:

\begin{alignat}{2} \tilde{f}(x) & \sim x \cdot Prob[X]\\ & \sim xf(x) \end{alignat}
\begin{alignat}{2} \tilde{f}(x) &= Prob[ \text{choosing x} ] \\ &= Prob[ \text{choosing x and at least one sample had value x} ] \\ &= Prob[ \text{choosing x} | \text{one sample had value x} ]Prob[ \text{one sample had value x} ] \\ &= xKf(x) \end{alignat}

K is some constant.

Normalizing pdf:

\int^\infty_0 \tilde{f}(x) dx = 1
K \int_0^\infty x f(x) dx = 1
K = \frac {1} {<X>}
\tilde{f}(x) = \frac {xf(x)} {<X>}

Distribution of Y

P[Y \le y | X = x] = \frac {y} {x}

This is uniform CDF, so the pdf:

\tilde f(y|x)=\frac {1}{x}

Now let's get \tilde f(y):

\begin{alignat}{2} \tilde{f}(y) &= \int^\infty_0 \tilde{f}(x,y)dx \\ &= \int^y_0 \tilde{f}(x,y)dx + \int_y^\infty \tilde{f}(x,y)dx \\ & Because \ Y \le X, \int^y_0 \tilde{f}(x,y)dx = 0 \\ &= \int^\infty_y \tilde{f}(y|x) \tilde{f}(x) dx \\ &= \int^\infty_y \frac {1} {x} ( \frac {x}{<X>} f(x))dx \\ &= \frac {1} {<X>} \int^\infty_y f(x) dx \\ &= \frac {1} {<X>} ( \int_0^\infty f(x)dx - \int_0^y f(x)dx) \\ &= \frac {1} {<X>} ( 1 - F(y)) \\ \end{alignat}

Use Laplace Transform:

Suppose

f\left(x\right) \sim F^*(s),

Then

F(x) \sim \frac{F^*(s)}{s}.

so

\,\tilde{f}^*(s)=\frac{1}{<X>}(\frac{1-F^*(s)}{s})
\,\frac{\part^n}{\part s^n}F^*(s)(s=0)= <X^n>

We may differentiate \tilde{f}^*(s) to obtain the moments of residual life. And by applying L'Hospital's rule, we get:

\,<Y^n>= \frac {1}{<X>} \frac {<X^{n+1}>}{n+1}

So,

\begin{alignat}{2} <Y> &= \frac {<X^2>}{2<X>} \\ &= \frac {<X>^2 + \sigma^2}{2<X>} \\ &= \frac {<X>}{2} + \frac{\sigma^2}{2<X>} \\ &\ge \frac {<X>}{2} \end{alignat}

x is exponential

<Y> = \frac{1}{2\lambda}+\frac{1}{2\lambda}=\frac{1}{\lambda}

Lecture_Voice_Recording

lecture_recording

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