Find the constant C in the following bivariate density.
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![(1) Find the constant C in the following bivariate density.
Scx² y? if x € (-1, 1) and y E (r², 1)
f(x, y) :
if otherwise.
The correct answer is
24/7
24
7
4/27
27/4
N/A
(Select One)
(2) Prove that the volume under the bivariate normal density is one. [Hint: by a substitution eliminate the constants a1, a2, b1, b2 from the problem.]
Here is the proof. Justify the equalities at the marked spots.
By the change of variable U =
(x – aj)/b1 and V = (y – a2)lb2, no loss of generality takes place if we assume that A1 = a2 =
O and that b1 = b2 = 1. Take 27 /1 – p² = (1/C). Then we have
%3D
00
00
-1
(4² +x² – 2pxy) } dy = C exp{ 2a-
- p?)
6² – 2pxy + p°x² + (x² – px)) } dy
exp
2(1 -
00
-1
@ Ce-12
(y – px)² } dy
2(1 – p2)
exp
-00
1
Ce-2 /27(1 – p²):
Then, (iii), explain why does the prove derivation prove that the volume under the bivariate normal density is one.
The correct answers for the three parts are:
Taylor expansion
Derivative in y
Integration by parts
Pulled e-x²/2 out of integral
None of the above
N/A
(i- Select One)
Taylor expansion
Derivative in y
Integrating a normal density
Integration by parts
None of the above
N/A
(ii- Select One)
Integrating a density gives 1
Taylor expansion
Derivative in y
Integration by parts
None of the above
N/A
(iii- Select One)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6d1bdda-b7a7-4c4f-bf66-52ea450cc4fa%2F63d5cb4a-7304-424b-834d-b98a96335c64%2Fl3jlhl_processed.png&w=3840&q=75)
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