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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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plz giv eme accurate soution thanks
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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