Applied Engineering Statistics | Questions-Answers

Applied Engineering Statistics
Problem 1:
Assume that the instrumental distribution g in importance
sampling with unstandardized weights is chosen such that ;
F(x)<M.g(x)
For all x and a suitable M ∈R ,where f is the target distribution.
Problem(1.a)
A random variable X drawn with the SIR algorithm has the
distribution that converges to f as m → ∞
Define w*(x)= f (x )
g ( x)
Let x1,...................,xm be i.i.d from g and consider an even B such that ,
P(Y∈ B∨X1,.........Xn)=∑
i=1
m
xi ∈ B ¿ w∗( xi)/∑
i=1
m
w∗( xi)¿
The law of large numbers shows that,
∑
i=1
m
¿ ( xi ∈ B ) w∗(Xi)→ E {∨ ( x ∈ B ) w∗( Xi)/∑
i=1
m
w∗( x )}
=∫
B
❑
w∗ ( x ) g ( x ) dx
And
Therefore,
∑
i=1
m
w∗( x ) → E {w∗( X ) }=∫ w∗( x ) g( x)dx=1
Hence M-1

It is clear that E that E[w*(x)]=E¿]=1
E[μis*]= 1
n ∑
i=1
n
E [h ( x ) w∗( xi ) ]=μ
Var[μ^* = 1
n2 ∑
i=1
n
var ( h ( xi ) w∗( xi ) ) =¿ 1
n ¿var(h(x)w(x)
This is the value that you are to reduce by a choice of w*(g)
You want f (x )
g ( x) to be bounded.
It is excellent to have w *(g) to be a heavier tail than f.
Avoid a rare draw from g getting a huge weight.
g should be nearly proportional to |h(x)f(x)| so that |h(x)f(x)/g(x) is
nearly a constant.
W* as defined are unstandardized weights by letting w(xi)=
w∗( xi)
∑
i=1
n
w∗(xi)
To obtain μ^=∑
i=1
n
h ( x ) w (xi)
Problem (1.b)
Let X1,X2,X3,........XN be identically distributed random variables
and h a fnction such that ;
Varg(w*(x).h(x)<∞ for all as N→∞.
Then,
TN=def 1
N ∑
i=1
N
h(xi)→pE(h(x))
= 1
N ∑
i=1
N
¿ ¿(i))w*(x)p→Ep(w*(x)h(x))

This implies that for all Ƹ >0
P(| 1
N ∑
I=1
N
h ¿ ¿(i))w*(x)-∫
x
❑
h ( x ) f ( x ) dx |≥ ∈ )
¿ ¿
You notice that P → 0as N→ ∞
The importance approach is upon the principle of exponential h(x)
with respect to density of f and can be expressed as ;
μ =∫ h ( x ) f ( x ) dx =∫ h( x) f ( x)
g (x).g(x) d(x)
Alternatively,
μ=
∫ h ( x ) f ( x ) dx
∫ f ( x)
g ( x ) dx
=
∫h( x ) f ( x )
g( x )
∫ f ( x )
g ( x ) . g ( x ) d (x)
.g(x) d(x)
Where g is the importance for sampling fnction and another
density to sample successfully.
This alternative form suggest that the approach to estimating
E[h(x)] is to draw X1,X2,........Xn i.i.d from g to use;
μ = 1
N ∑
i=1
n
h ( xi ) w∗( xi )
¿ ¿
Where w*(xI)= f ( xi)
g (xi) is the importance ratio.
Problem 2
> qnorm(0.95,mean=0,sd=1)
[1] 1.644854

> x<-rnorm(1000,mean=0,sd=1)
> hist(x,probability=TRUE)
>
> qnorm(0.95,mean=1,sd=1.414)
[1] 3.325823
> x<-rnorm(1000,mean=1,sd=1.42)
> hist(x,probability=TRUE)

>
>
The Weighted Variance
> x<-c(1:1000)
> w<-rpois(1000,1000)+1
> y<-x[rep(seq_along(x),w)]
> var(y)

[1] 83330
>
The Weighted Mean
> x<-c(1:1000)
> w<-rpois(1000,1000)+1
> y<-x[rep(seq_along(x),w)]
> mean(y)
[1] 500.6345
>
1:1000 is a set of numbers from 1 to 1000.
Wt- is some arbitrary weights .
Var(y) –is the weighted variance
Mean(y) – is the weighted mean.

The histogram from the weighted means and the weighted
variance is more uniformly and evenly distributed.
Problem(3a)

Consider P to be doubly stochastic matrix associated with the
transition probabilities of the Markov chain with N opinions.
The limiting-state probabilities are given by ⫪i= 1
N ;i=1,2,....,N
Proof:
The limiting-state probabilities are to satisfy the condition ;
⫪i=∑
k
❑
⫪kPkj
Substitute ⫪i= 1
N ,i=1,2,......,N
The result is as follows;
1
N = 1
N ∑
kj
❑→1=∑
k
Pkj
This confirms the validity of the theorem.
The condition ⫪=⫪P is satisfied by ⫪i= 1
N
Where ⫪=⫪P is a limiting state probabilities needed for confirmation.
Each column j of P have a simulation of 1 ,supposing that the limiting –state
probabilities are given by ⫪i= 1
N
This confirms P to be a doubly stochastic ;
Positive Negative Worse
Positive 1
3
1
3
1
3
P=
Negative 1
3
1
3
1
3