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Problem 1
Solution
Suppose X maps W to R ,meaning the w ∈ Rand also X (w)∈ R
Now Y(w)=X |w|3 also maps¿ Y :W → R
Now themeasure for cumulative distribution function will be as below
F ( x )=P ( X ≤ x )
We also consider y> 0
FY ( y )=P (Y ≤ y )
¿ P (|x|3 ≤ y )
¿ P(− 3
√ y )≤ X ≤− 3
√ y
Now
¿ FX (−3
√ y )−FX (−3
√ y )+ P(X =−3
√ y)
since X is a random variable ,the |x|3 is alsoa random variable on the same space .
Problem 2
Part a
Solution
Using the formula∫∫
−∞
∞
f ( x , y ) dydx
E ( x , y ) =∫
0
2
∫
0
2
c ( y2 + xy ) dydx
∫
0
2
∫
0
2
c ( y2 + xy ) dydx=1

∫
0
2
c [ y3
3 + x y2
2 ] 3
0 dx=c ¿ ¿
¿ c [ 9 x+ 9
4 x2
] 2
0=c [ 18+ 9 ] =1
27 c=1
therefore , c= 1
27
Part b
Solution
Given that ∫∫
−∞
∞
f ( x , y ) dydx
we will thenfind theindividual marginal functions of X∧Y .
now the marginal density for X will be given as
f ( x ) =∫
−∞
∞
f ( x , y ) dy
¿∫
0
3
y2+ xy
27 dy
¿ 1
27 [ y3
3 + x y2
2 ] 3
0= 1
27 [ 9+ 9
2 x ]
¿ 1
27 ( 9 )+ 1
27 ( 9
2 )x= 1
3 + 1
6 x
¿ 2+ x
6
Therefore , f ( x )= 2+ x
6
Also,
f ( y ) =∫
−∞
∞
f ( x , y ) dx

¿ 1
27 ∫
0
2
f ( y2+ xy ) dx
¿ 1
27 [ x y2+ x2 y
2 ] 2
0= 1
27 [ 2 y2+2 y ]
Therefore , f ( y ) = 2 y2+ 2 y
27
¿ the above obtained solutions , we can observe that f ( x , y ) ≠ f ( x ) f ( y )
Now we can confidently concludethat X ∧Y are not independent .
We now want ¿ calculte the xpected values ;
Expectedd values of 3 X −2 y is given by E [ 3 X −2Y ] =3 E [ X ] −2 E [ Y ] ... ... . A
Expected values of X will be givenby
E [ X ] =∫
−∞
∞
xf ( x ) dx
¿∫
0
2
1
6 x ( 2+ x ) dx= 1
6 ∫
0
2
2 x + x2 dx
¿ 1
6 [x2 + x3
3 ]2
0= 1
6 (4+ 8
3 )=10
9
Therefore , E ( x ) =10
9
Also ,
E [ Y ] =∫
−∞
∞
yf ( x , y ) dy
¿∫
0
3
y ( 2 y2 + y
27 )dy = 2
27 ∫
0
3
( y3 + y2 ) dy
¿ 2
27 [ y4
4 + y3
3 ]3
0= 2
27 [ 81
4 + 27
3 ]=13
6
Therefore , E [ Y ] =13
6

Now ¿ our definition ¿ equation A
E [ 3 X−2 Y ]=3 ( 10
9 )−2 ( 13
6 )=10
3 − 13
3 =−1
Therefoe , E [ 3 X−2 Y ] =−1
Covariance
We wil consider a general formula for the covariance
Var [ X +Y ] =E [ ( X +Y ) 2 ] − [ E [ X +Y ] ] 2
Therefore , Var ( ax+ by ) =E [ ( ax +by ) 2 ] − [ E [ ax +by ] ] 2
¿ E [ a2 x2+ b2 y2 +2 abxy ]−¿
¿ a2 E [ x2 ]+b2 E [ Y 2 ] +2 ab [ XY ]−a2 x2−b2 y2−2 ab x y
Therefore , Var ( ax+ by )=a2 σ2 x+ b2 σ2 y2+2 abCxy ... ... ... Equation B
Now we need ¿ find the σ 2 x ,σ 2 y ∧Cxy
E [ x2 ] =∫
−∞
∞
x2 f ( x ) dx=∫
0
2
x2 ( 2+ x
6 ¿¿) dx ¿ ¿
¿ 1
6 ∫
0
2
( 2 x2+ x3 ) dx= 1
6 ( 16
3 + 4 ) = 8
9 + 2
3 = 14
9
E [ x2 ]=14
9
Var ( x )=σ2 x=E [ x2 ]− [ E [ x ] ]2
¿ ( 14
9 )− ( 10
9 )
2
¿ 14
9 − 100
81 =26
81
Hence , σ2 x=0.32099
E [ Y 2 ] =∫
−∞
∞
y2 f ( x ) dy=∫
0
3 y2 ( 2 y2 +2 y )
27 dy