Question

Help me about mathematical statistics please? I really need help for preparing my midterm in mathematical statistics. Thank you very much for your help.

Answer 1

Given `f_(X,Y)(x,y)={(4x^2y+2y^5",", 0 <= x <= 1", " 0 <= y <= 1), (0",", "otherwise")]`

Let `A` be the set of all possible values of `X`. Then `A=[0,1].`
Let `B` be the set of all possible values of `Y`. Then `B=[0,1].`
The support is the 2-D region `{(x,y) | 0<=x<=1, 0<=y<=1}`.

a) `E(X)`

When `X` and `Y` are jointly distributed, we have the general formula for the expected value of a function of `X` and `Y`:

`E[g(X,Y)] = int_A int_B g(x,y)f_(X,Y)(x,y)" "dy" "dx`

Of course, the function `g(X,Y)` can be whatever we choose. If we choose `g(X,Y)=X`, we get

`E[X] = int_A int_B x" "f_(X,Y)(x,y)" "dy" "dx`

For this question:

`E(X) = int_(x=0)^1int_(y=0)^1x(4x^2y+2y^5)" "dy" "dx`

`color(white)(E(X)) = int_(0)^1 x int_(0)^1(4x^2y+2y^5)" "dy" "dx`

`color(white)(E(X)) = int_(0)^1 x [2x^2y^2+1/3 y^6]_(y=0)^1" "dx`

`color(white)(E(X)) = int_(0)^1 x (2x^2+1/3)" "dx`

`color(white)(E(X)) = int_(0)^1 (2x^3+1/3 x)" "dx`

`color(white)(E(X)) = [1/2 x^4+1/6 x^2]_(x=0)^1`

`color(white)(E(X)) = 1/2 + 1/6`

`color(white)(E(X)) = 2/3`

b) `E(Y)`

I'll leave the calculation of `E(Y)` as an exercise. Just choose `g(X,Y)=Y.` You should get `E(Y) = 46/63.`

c) `E(3X+7Y)`

Using the rules of expectation, we get

`E(3X+7Y)=E(3X) + E(7Y)`

`color(white)(E(3X+7Y))=3E(X) + 7E(Y)`

`color(white)(E(3X+7Y))=3(2/3) + 7(46/63)`

`color(white)(E(3X+7Y))=2+46/9`

`color(white)(E(3X+7Y))=64/9`

d) `"Var"(X)`

Recall by the definition of variance:

`"Var"(X) = E{[X-E(X)]^2} = E(X^2)- [E(X)]^2`

We already know `E(X)`, so we just need `E(X^2).` This is found by letting `g(X,Y) = X^2:`

`E(X^2) = int_(x=0)^1int_(y=0)^1x^2(4x^2y+2y^5)" "dy" "dx`

`color(white)(E(X^2)) = int_(0)^1 x^2 int_(0)^1(4x^2y+2y^5)" "dy" "dx`

`color(white)(E(X^2)) = int_(0)^1 x^2 [2x^2y^2+1/3 y^6]_(y=0)^1" "dx`

`color(white)(E(X^2)) = int_(0)^1 x^2 (2x^2+1/3)" "dx`

`color(white)(E(X^2)) = int_(0)^1 (2x^4+1/3 x^2)" "dx`

`color(white)(E(X^2)) = [2/5 x^5+1/9 x^3]_(x=0)^1`

`color(white)(E(X^2)) = 2/5 + 1/9`

`color(white)(E(X^2)) = 23/45`

Thus,

`"Var"(X) = E(X^2)- [E(X)]^2`

`color(white)("Var"(X)) = 23/45- (2/3)^2`

`color(white)("Var"(X)) = 23/45- 4/9`

`color(white)("Var"(X)) = 1/15`

e) `"Var"(Y)`

Once again, I'll leave this one as practice. You should get `"Var"(Y)=797/15876~~0.0502.`

f) `E(XY)`

Same as before, but with `g(X,Y)=XY`. You should get `E(XY)=10/21.`

g) `E(XY+14)`

We use the properties of expectation again to get

`E(XY+14)=E(XY) + E(14)`

`color(white)(E(XY+14))=10/21 + 14`

`color(white)(E(XY+14))=304/21`

h) `"P"(X+Y<=1)`

We're looking for the probability of being in the part of the support that satisfies `x+y<=1`. So we need to integrate `f_(X,Y)(x,y)` over this partial region. The added restriction is:

`x+y<=1`
`=>" "y<=1-x`

Combining this with the given restrictions `x>=0` and `y>=0,` our integration bounds are the `x`-axis, the `y`-axis, and `y=1-x.`

We can picture covering this region as follows: as `x` goes from `0` to `1,` `y` will go from `0` to `1-x.`

Thus, our probability is

`"P"(X+Y<=1)= int_(x=0)^1int_(y=0)^(1-x)f_(X,Y)(x,y)" "dy" "dx`

`color(white)("P"(X+Y<=1))= int_(0)^1int_(0)^(1-x)(4x^2y+2y^5)" "dy" "dx`

`color(white)("P"(X+Y<=1))= int_(0)^1 [2x^2y^2]_(y=0)^(1-x)" "dx`

`color(white)("P"(X+Y<=1))= int_(0)^1 2x^2(1-x)^2" "dx`

`color(white)("P"(X+Y<=1))= int_(0)^1 (2x^2-4x^3+2x^4)" "dx`

`color(white)("P"(X+Y<=1))= [2/3x^3-x^4+2/5x^5]_(x=0)^1`

`color(white)("P"(X+Y<=1))= 2/3-1+2/5`

`color(white)("P"(X+Y<=1))= 1/15`

Good luck on your midterm!