# KarlRosaen

## Problem

Let

$$f_{X,Y}(x,y) = \begin{cases} c(x+y) & 0 \leq x \leq 1, 0 \leq y \leq 2 \\ 0 & \text{otherwise} \end{cases}$$

Where $c$ is a constant. Find $V(2X - 3Y + 8)$.

## Solution

Solving this problem involves a few steps:

1. Finding the constant c by noting that integrating the joint distribution must equal 1
2. Finding the marginal distributions for X and Y
3. Finding E(X), E(Y), V(X), V(Y) and Cov(X,Y)
4. Breaking down V(2X - 3Y +8) into components of the above to make the final calculation

### 1. Finding C

$\int_0^1\int_0^2 c(x+y) dy dx = c \int_0^1 xy + \frac{y^2}{2} \rvert_{y=0}^{y=2} dx = c \int_0^1 2x + 2 dx = c(x^2 + 2x \rvert_{x=0}^{x=1} = c(1+2) = 3c$

Probability distributions must integrate over all possible values to equal 1, so $3c = 1$ and $c = \frac{1}{3}$

### 2. Finding marginals

Finding the marginals is a matter of integrating out the other term.

$f_X(x) = \frac{1}{3} \int_{0}^2 x + y dy = \frac{1}{3} (xy + \frac{y^2}{2} \rvert_{y=0}^{y=2}) = \frac{1}{3} (2x + 2) = \frac{2x}{3} + \frac{2}{3}$

$f_Y(y) = \frac{1}{3} \int_{0}^1 x + y dx = \frac{1}{3} (\frac{x^2}{2} + xy \rvert_{x=0}{x=1}) = \frac{1}{3}(\frac{1}{2} + y) = \frac{y}{3} + \frac{1}{6}$

### 3. Finding all the expectations and (co)variances

$E(X) = \int x f_X(x) dx = \int_0^1 \frac{2}{3}x^2 + \frac{2}{3}x = \frac{2}{9}x^3 + \frac{1}{3}x^2 \rvert_{x=0}^{x=1} = \frac{2}{9} + \frac{1}{3} = \frac{5}{9}$

$E(X^2) = \int x^2 f_X(x) dx = \int_0^1 \frac{2}{3}x^3 + \frac{2}{3}x^2 = \frac{1}{6}x^4 + \frac{2}{9}x^3 \rvert_{x=0}{x=1} = \frac{1}{6}+\frac{2}{9} = \frac{3}{18} + \frac{4}{18} = \frac{7}{18}$

$E(Y) = \int y f_Y(y) dx =\int_0^2 \frac{y^2}{3} + \frac{y}{6} dy = \frac{y^3}{9} + \frac{y^2}{12} \rvert_{0}^2 = \frac{8}{9} + \frac{4}{12} = \frac{16}{18} + \frac{6}{18} = \frac{22}{18} = \frac{11}{9}$

$E(Y^2) = \int y^2 f_Y(y) dx =\int_0^2 \frac{y^3}{3} + \frac{y^2}{6} dy = \frac{y^4}{12} + \frac{y^3}{18} \rvert_0^2 = \frac{16}{12} + \frac{8}{18} = \frac{4}{3} + \frac{4}{9} = \frac{12}{9} + \frac{4}{9} = \frac{16}{9}$

$V(X) = E(X^2) - E(X)^2 = \frac{7}{18} - \frac{25}{81} = \frac{63}{162} - \frac{50}{162} = \frac{13}{162}$

$V(Y) = E(Y^2) - E(Y)^2 = \frac{16}{9} - \frac{121}{81} = \frac{144}{81} - \frac{121}{81} = \frac{23}{81}$

$E(XY) = \frac{1}{3} \int \int xy(x+y)dx dy = \frac{1}{3} \int_0^2 \int_0^1 x^2y + xy^2 dx dy = \frac{1}{3} \int_0^2 (\frac{x^3y}{3} + \frac{x^2y^2}{2} \rvert_{x=0}^{x=1}) dy = \frac{1}{3} \int_0^2 \frac{y}{3} + \frac{y^2}{2} dy = \frac{1}{3} (\frac{y^2}{6} + \frac{y^3}{6} \rvert_0^2) = \frac{1}{3}(\frac{4}{6} + \frac{8}{6}) = \frac{2}{3}$

$Cov(X,Y) = E(XY) - E(X)E(Y) = \frac{2}{3} - \frac{5}{9}\frac{11}{9} = \frac{2}{3} - \frac{55}{81} = \frac{54}{81} - \frac{55}{81} = -\frac{1}{81}$

### 4. Putting it all together

Phew. Now we have everything we need to calculate $V(2X - 3Y + 8)$.

First, $V(2X - 3Y + 8) = V(2X - 3Y)$ as the constant has no variance. Applying the formula for linear combinations of variance,

$V(2X-3Y) = 2^2 V(X) + 3^2V(Y) - 2 \times 2 \times 3 Cov(X,Y) = 4V(X) + 9V(Y) - 12 Cov(X,Y) \\ = 4\frac{13}{162} + 9\frac{23}{81} - 12\frac{-1}{81} = \frac{26}{81} + \frac{207}{81} + \frac{12}{81} = \frac{245}{81}$

## Sources

• 36-700 hw4 problem 3