Difference between revisions of "EBook Problems MultivariateNormal"

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(→‎EBook Problems Set - Mutivariate Normal Distribution)
(→‎EBook Problems Set - Mutivariate Normal Distribution)
Line 27: Line 27:
  
 
Let ''T2'' be Person2’s travel time in (hours);  
 
Let ''T2'' be Person2’s travel time in (hours);  
 +
 
<math>
 
<math>
−
T2 = T_3 + T_4
+
T2 = T_3 + T_4 \,
 
</math>
 
</math>
 
with
 
with
 
<math>
 
<math>
−
\mu_{T2}=5+4=9(hours),  
+
\mu_{T2}=5+4=9(hours) \,  
 
</math>
 
</math>
 +
 
and
 
and
  
−
<math>\sigma_{T2}=[3^2+1^2+(2)(0.8)(3)(1)]^{1/2}=3.847076812 (hours)</math>
+
<math>\sigma_{T2}=[3^2+1^2+(2)(0.8)(3)(1)]^{1/2}=3.847076812 (hours) \,</math>
  
 
Hence
 
Hence
Line 52: Line 54:
  
 
<math>
 
<math>
−
\mu_{T1}=6+4=10(hours),
+
\mu_{T1}=6+4=10(hours), \,
 
</math>
 
</math>
 +
 
and
 
and
  
 
<math>
 
<math>
−
\sigma_{T1}=[2^2+1^2]^{1/2}=\sqrt{5}
+
\sigma_{T1}=[2^2+1^2]^{1/2}=\sqrt{5} \,
 
</math>
 
</math>
  
Line 63: Line 66:
  
 
<math>
 
<math>
−
P(T2-T1>1)=P(T1-T2+1<0),
+
P(T2-T1>1)=P(T1-T2+1<0), \,
 
</math>
 
</math>
  
Line 69: Line 72:
  
 
<math>
 
<math>
−
\mu_R=\mu_{T1}-\mu_{T2}+1=10-9+1=2,
+
\mu_R=\mu_{T1}-\mu_{T2}+1=10-9+1=2, \,
 
</math>
 
</math>
  
 
<math>
 
<math>
−
\sigma_R=[\sigma_{T1}^2+\sigma_{T2}^2]^{1/2}=(5+14.8)^{1/2}=\sqrt{19.8},
+
\sigma_R=[\sigma_{T1}^2+\sigma_{T2}^2]^{1/2}=(5+14.8)^{1/2}=\sqrt{19.8}, \,
 
</math>
 
</math>
  
Line 79: Line 82:
  
 
<math>
 
<math>
−
P(R<0)=P(\frac{R-\mu_R}{\sigma_R}<\frac{0-2}{\sqrt{19.8}})
+
P(R<0)=P(\frac{R-\mu_R}{\sigma_R}<\frac{0-2}{\sqrt{19.8}}) \,
 
</math>
 
</math>
  
Line 85: Line 88:
  
 
<math>
 
<math>
−
=\Phi (-0.44946657)
+
=\Phi (-0.44946657) \,
 
</math>
 
</math>
  
 
<math>
 
<math>
−
\cong 0.327
+
\cong 0.327 \,
 
</math>
 
</math>
 
}}
 
}}
Line 95: Line 98:
 
Since the A<math>\rightarrow</math>C<math>\rightarrow</math>D route has a smaller expected travel time of <math>\mu_{T2}=9</math> hours as compared to the upper (with expected travel time  <math>=\mu_{T1}=10 hours)</math>, one should take the '''lower''' route to minimized expected travel time from A to D.
 
Since the A<math>\rightarrow</math>C<math>\rightarrow</math>D route has a smaller expected travel time of <math>\mu_{T2}=9</math> hours as compared to the upper (with expected travel time  <math>=\mu_{T1}=10 hours)</math>, one should take the '''lower''' route to minimized expected travel time from A to D.
 
}}
 
}}
 +
 +
===Problem 2===
 +
The settlement (in cm) of a structure shown in the following figure may be evaluated from
 +
 +
<math>
 +
S=0.3A+0.2B+0.1C
 +
</math>
 +
 +
[[Image:Problem 2.jpg|250x250px|center]]
 +
 +
where A, B, and C are respectively the thickness (in m) of the three layers of soil as shown. Suppose A, B, and C are modeled as independent normal random variables as
 +
 +
A ~ N(5,1)
 +
 +
B ~ N(8,2)
 +
 +
C ~ N(7,1)
 +
 +
 +
''(a) Determine the probability that the settlement will exceed 4 cm.
 +
 +
''(b) If the total thickness of the three layers is known exactly as 20 m; and furthermore, thicknesses A and B are correlated with correlation coefficient equal to 0.5, determine the probability that the settlement will exceed 4 cm.
 +
 +
{{hidden|Answer for part (a)|
 +
To calculate probability, we first need to have the PDF of S. As a linear combination of three normal variables, S itself is normal, with parameters
 +
 +
<math>
 +
\mu_s=0.3 X 5 + 0.2 X 8 + 0.1 X 7 = 3.8(cm) \,
 +
</math>
 +
 +
<math>
 +
\sigma_s=\sqrt{0.3^2 1^2+0.2^2 2^2+0.1^2 1^2} \cong 0.51 (cm) \,
 +
</math>
 +
 +
Hence
 +
 +
<math>
 +
P(S>4cm) = P(\frac{S-\mu_s}{\sigma_s}>\frac{4-3.8}{0.509901951}) \,
 +
</math>
 +
 +
<math>
 +
= 1-\Phi(0.39223227)=1-0.65255665 \,
 +
</math>
 +
 +
<math>
 +
\cong0.347 \,
 +
</math>
 +
}}
 +
{{hidden|Answer for part (b)|
 +
Now that we have a constraint <math>A+B+C=20m</math>, these variables are no longer all independent, for example, we have
 +
 +
<math>
 +
C=20-A-B \,
 +
</math>
 +
 +
Hence
 +
 +
<math>
 +
S=0.3A+0.2B+0.1(20-A-B) \,
 +
</math>
 +
 +
<math>
 +
\Rightarrow S=0.2A+0.1B+2 \,</math>
 +
 +
with
 +
 +
<math>
 +
\rho=0.5 \,
 +
</math>
 +
 +
Thus
 +
 +
<math>
 +
\mu_s=0.2 x 5+0.1 x 8+2=3.8(cm) \,</math>
 +
 +
as before, and
 +
 +
<math>
 +
\sigma_s=\sqrt{0.2^2 1^2+0.1^2 2^2+2 x 0.5 x 0.1 x 1 x 2} \,
 +
</math>
 +
 +
<math>
 +
= \sqrt{0.12} \cong 0.346(cm) \,
 +
</math>
 +
 +
Hence
 +
 +
<math>
 +
P(S>4cm)=1-\Phi (\frac{4-3.8}{\sqrt{0.12}}) \,
 +
</math>
 +
 +
<math>
 +
=1-\Phi(0.57735027)=1-0.718148613 \cong 0.282 \,
 +
</math>
 +
}}
 +
 +
 +
 +
 
<hr>
 
<hr>
 
* [[EBook | Back to Ebook]]
 
* [[EBook | Back to Ebook]]

Revision as of 15:21, 11 January 2011

EBook Problems Set - Mutivariate Normal Distribution

Problem 1

Person1 and Person2 are travelling from point A to point D, but there are different routes to get from A to D. Person1 decides to take the A->B->D route, whereas Person2 takes the A->C->D route.

Problem 1.jpg

The travel times (in hours) between each pair of points indicated are normally distributed as follows:

T1 ~ N (6, 2)

T2 ~ N (4, 1)

T3 ~ N (5, 3)

T4 ~ N (4, 1)

Explain why these times are stochastic (and not exact or deterministic)? Although the travel times here generally can be assumed statistically independent, T3 and T4 are dependent with correlation coefficient 0.8.


(a) What is the probability that Person2 will not arrive at point D within 10 hours?

(b) What is the probability that Person1 will arrive at point D earlier than Person2 by at least one hour?

(c) Which route (A\(\rightarrow\)B\(\rightarrow\)D or A\(\rightarrow\)C\(\rightarrow\)D) should be taken if one wishes to minimize the expected travel time from A to D? Explain.

Answer for part (a)

{\sigma_{T2}}\leq\frac{10-9}{3.847076812})

=1-\Phi(0.259937622)=1-0.602543999

\cong 0.397 \) }}

Answer for part (b)

) \, \)

Hence

\( =\Phi (-0.44946657) \, \)

\( \cong 0.327 \, \) }}

Answer for part (c)


Problem 2

The settlement (in cm) of a structure shown in the following figure may be evaluated from

\( S=0.3A+0.2B+0.1C \)

Problem 2.jpg

where A, B, and C are respectively the thickness (in m) of the three layers of soil as shown. Suppose A, B, and C are modeled as independent normal random variables as

A ~ N(5,1)

B ~ N(8,2)

C ~ N(7,1)


(a) Determine the probability that the settlement will exceed 4 cm.

(b) If the total thickness of the three layers is known exactly as 20 m; and furthermore, thicknesses A and B are correlated with correlation coefficient equal to 0.5, determine the probability that the settlement will exceed 4 cm.

Answer for part (a)
Answer for part (b)

) \, \)

\( =1-\Phi(0.57735027)=1-0.718148613 \cong 0.282 \, \) }}



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