Steel Grain Size Yield Strength Simulation
This set of notes was developed as a result of an in-class example discussed on 2/14/2012. In that
example the following random variables were addressed:
X = The act of measuring the overall grain size in a specimen of steel
Y = The act of measuring the yield strength for that specimen.
The following assumptions were made:
(A1) X ~ N ( X  10, X  3) ; Y ~ N (Y  100, Y  10)
(A2) rXY  0.3
Remark 1. It was remarked by a student in material science that, in fact, rXY is closer to zero for smaller
values of x.
In order to simulate measurements of {( X k , Yk )}nk 1 for sample size n=30, using Matlab, it was necessary
to specify
 
 XY   X 
 Y 
and
 2
 XY   X
 XY
 XY 
.
 Y2 

Since rXY   XY /( X  Y ) , we obtained  XY  rXY ( X  Y )  0.3(3)(10)  9 . The Matlab command
xy = mvnrnd(  XY , 
 , 30) was used to generate a single simulation for sample size 30.
Assumption (A2) reflects the assumption that the following linear prediction model is appropriate:

Y  mX  b .
(1)
The model parameters m and b need to be estimated from the data. To this end, we demanded that the
following two conditions should hold:
(C1) Y  Y . In words, the estimator (1) is required to be an unbiased estimator of Y.


(C1) Let W  Y  Y denote the prediction error. Then we require  W X  0 .
From these two conditions we have the following two equations in the two unknowns, m and b:
Y  Y


E (Y )  E (mX  b)  m X  b  Y

(2a)

 W X  Cov(W , X )  Cov(Y  Y , X )  Cov(Y , X )  Cov(Y , X )
  XY  Cov(mX  b, X )   XY  mCov( X , X )   XY  m X2  0
(2b)
Hence, the solutions for the unknown parameters m and b are: m   XY /  X2 and b  Y  m X .
It follows that the natural estimators of these parameters are:
 

 

m   XY /  X2 and b  Y  m X
(3)
where the hat above a parameter denotes the associated moment estimator of it. For example, since the



parameter  X  E(X ) , then  X  E ( X ) 
1 n
 X k . While the expressions in (3) may not appear
n k 1
familiar to some readers, they are, in fact, the expressions for the Least Squares (LS) estimators of the
associated parameters. This fact is addressed in the Chapter 5 notes.
The Intent of This Set of Notes
The intent of this set of notes is to demonstrate how simulations can be used to accommodate
Remark 1 above. Questions that will be answered include:
(Q1): How can one develop more realistic simulations that reflect Remark 1?
(Q2): In the case of those more realistic simulations, what happens when one uses the linear
model (1), without the knowledge of the simulation structure that generated the data?
In answering these two questions, two very important concepts will be demonstrated:
Concept 1: Simulation of jointly normal random variables, (X,Y), when the correlation
coefficient is not a constant; rather it is a function of x.
Concept 2: The power of using simulations to investigate the statistics of parameter estimators
when mathematical methods become difficult, if not intractable.
A Model for rXY(x) : In view of Remark 1, we desire a model for rXY (x) that satisfies the
condition that as the grain size, x, gets smaller, so does the magnitude of rXY (x) . The model that
we will use for this purpose is:
 0.7 for x   X   X
.
rXY ( x)  
[ 0.5(  X  X  x )]
for x   X   X
 0.7e
(4)
This model is plotted in Figure 1.
0
-0.1
-0.2
r(x)
-0.3
-0.4
-0.5
-0.6
-0.7
0
2
4
6
8
10
12
Overall Grain Size (x)
14
16
18
20
Figure 1. Plot of rXY (x) given by (4) for  X  10 and  X  3 .
Results for a single simulation with sample size n = 30 is given in Figure 2.
Scatter Plot of Last Simualtion
130
Yield Strength (y)
120
110
100
90
80
70
2
4
6
8
10
Overall Grain Size (x)
12
14
16
Figure 2(a) Scatter plot of x-y.
Plot of r(k) for last simulation
0
-0.1
-0.2
r(k)
-0.3
-0.4
-0.5
-0.6
-0.7
0
5
10
15
20
Specimen Number (k)
Figure 2(b) Values of rXY (x) for the simulation in Figure 2(a).
25
30
The results for 1000 simulations with n=30 are shown in Figure 3.
Histogram of Slope estimate
300
250
200
150
100
50
0
-5
-4
-3
-2
-1
0
1
Histogram of Intercept estimate
350
300
250
200
150
100
50
0
80
90
100
110
120
130
140
150
Scatter Plot of Slope vs Intercept estimate
150
140
Intercept Estimate
130
120
110
100
90
80
-5
-4
-3
-2
Slope Estimate
-1
0
1
Histogram of Sample Average Correlation Coefficient
250
200
150
100
50
0
-0.45
-0.4
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1


Figure 3. Estimated shapes of the pdf’s for m and b (top two plots), a scatter plot indicating the
joint pdf structure (third), and the estimated shape of the pdf for the average of the correlation
coefficient, rXY (X ) .
Your Notes of the In-Class Discussion of the Above Results:
1.
APPENDIX Matlab Codes
% PROGRAM NAME: corrxy.m
x - 0:.1:20;
nx = length(x);
r = -0.7*ones(1,nx);
for k = 1:nx
if x(k)<13
r(k) = -.7*exp(-.5*(13 - x(k)));
end
end
figure(1)
plot(x,r)
xlabel('Overall Grain Size (x)')
ylabel('r(x)')
grid
% PROGRAM NAME: steel.m
% This code simulates measurements of (X,Y)
% X = grain size SX = [0,inf)
% Y = Yield strength SY = [0,inf)
% ASSUMPTIONS:
% (A1): E[X ; Y] = [ 10 100]
mux = 10; muy = 100;
% (A2): Var(X)=3^2 & Var(Y)=10^2
sigx = 3; sigy = 10;
% (A3): Corr(X,Y) = rxy = rmax*exp(([1- 2x)) for x < mux+sigx
%
= 0.7 for x > mux+sigx
rmax = -.7; msx = mux + sigx;
n = 30; % Sample Size
nsim = 1000; % Number of simulations
%===================================
M = zeros(1,nsim); B = M; x = zeros(n,1); y = x;
r=x; R=[];
for ksim = 1:nsim
r=rmax*ones(n,1);
for k = 1:n
x(k) = normrnd(mux,sigx);
if x(k) < msx
r(k) = rmax*exp(-0.5*(msx-x(k)));
end
m = r(k)*(sigy/sigx);
b = muy - m*mux;
sigu = (sigy^2 - m^2*sigx^1)^0.5;
u = normrnd(0,sigu);
y(k) = m*x(k) + b + u;
end
R = [R r];
mx=mean(x); my=mean(y); Cxy=cov(x,y); cxy=Cxy(1,2); vx=Cxy(1,1);
M(ksim) = cxy/vx;
B(ksim) = my - M(ksim)*mx;
end
figure(1)
plot(x,y,'*')
xlabel('Overall Grain Size (x)')
ylabel('Yield Strength (y)')
title('Scatter Plot of Last Simualtion')
grid 'minor'
pause
figure(2)
plot(r,'*')
xlabel('Specimen Number (k)')
ylabel('r(k)')
title('Plot of r(k) for last simulation')
grid 'minor'
pause
figure(3)
hist(M)
title('Histogram of Slope estimate')
pause
figure(4)
hist(B)
title('Histogram of Intercept estimate')
pause
figure(5)
plot(M,B,'*')
title('Scatter Plot of Slope vs Intercept estimate')
xlabel('Slope Estimate')
ylabel('Intercept Estimate')
pause
figure(6)
Rm = mean(R);
hist(Rm)
title('Histogram of Sample Average Correlation Coefficient')