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Assignment.2 (Lessons 24-27)
Question: Marks: 5++5+5=15
A joint function of (X,Y) is given by the following equation:
(a) Show that given fulfills the conditions of a p.d.f.
(b) Compute the Marginal Probability density function of variable X
(c) Calculate the Expected value of X i.e, E(X).
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is it correct or not related to the assignment
how it solved in last three lines and middle line?
Need assistance please or share ideas
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Asa, Plz can anyone tell from where i may get the math type software?
yes u can find the application in vu lms download section mathtype 6.9 full rar etc ye name likha ho ga to wahan se download ker k install ker len us k sath key b ho g
agr smj na aye to puch lena
I installed 30 days trail version from google not found from lms?
download section k second page pe likha he ye MathType 6 full.rar
es pe click ker k download ker lo ok he gor se dekho
Q1) A joint function of (X,Y) is given by the following equation:
(a) Show that given fulfills the conditions of a p.d.f.
(b) Compute the Marginal Probability density function of variable X
(c) Calculate the Expected value of X i.e, E(X).
Solution a :
4xy^{2} / 3^{ } +4xy^{2} )dydx =1
(4xy^{2 } /6 + 4xy^{3} /3 )^{1}_{0 } dx =1
(2xy^{2 } /3 + 4xy^{3} /3 )^{1}_{0 } dx =1
{2x / 3 [ (1)^{2 } - (0)^{2}] +4x/3 [(1)^{3} – (0)^{3} ]} dx =1
(2x/ 3 + 4x/3) dx =1
( 2/3 (x^{2} /2) + 4/3 (x^{2} /2))^{1}_{0 } =1
(x^{2 } / 3 + 2/3x^{2})^{1}_{0 } = 1
1/ 3 [ (1)^{2} –(0)^{2}] + 2/3[ (1)^{2}+(0)^{2}] =1
by simplifying
3/3 =1
1=1 Hence F(x,y) is p.d.f
(b) g(x) =f(x,y)dy
4xy(1/3+y)dy
(4xy/3+4xy^{2})dy
Opening integration
[4 x/ 3 (y^{2} / 2) + 4x (y^{3} / 3)]^{1}_{0}
_{ }
2x/3[(1)^{2 } - (0)^{2}] + 4x / 3 [(1)^{3} – (0)^{3}]
g(x) = 2x / 3 + 4x/3 marginal probability of x
(c) E(x) = xg(x)dy
2x/3+ 4x / 3 )dx
2x^{2}/3+ 4x ^{2}/ 3 )dx
[2/3 (x^{3}/3)+ 4/3 (x^{3} /3)]^{1}_{0}
_{ }
simplifying
2 /9 [(1)^{3} – (0)^{3}] + 4/9 [(1)^{3 } - (0)^{3}]
2 / 9 + 4 / 9
6 / 9
E (x) = 2 /3
sorry for maths type
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