Question

The random variables X and Y have a joint PDF of the form f X, Y (x,y)=c* exp\ - 1 2 (4 x^ 2 -8 x+y^ 2 -6 y+13)\
E[X] =

Var (X) =

E[Y] =

Var (Y) =

Answer 1

To solve for the expected values \(E[X]\), \(E[Y]\) and variances \(Var(X)\), \(Var(Y)\) for the given joint PDF \(f_{X,Y}(x,y) = c \cdot \exp\left(-\frac{1}{2}(4x^2 - 8x + y^2 - 6y + 13)\right)\), we first need to identify the values of \(c\) and rewrite the expression in a more manageable form by completing the square for the terms involving \(x\) and \(y\).

### Step 1: Completing the square for \(x\)

For terms involving \(x\):
\[4x^2 - 8x = 4\left(x^2 - 2x\right)\]
Completing the square:
\[x^2 - 2x = (x - 1)^2 - 1\]
Therefore:
\[4(x^2 - 2x) = 4\left((x - 1)^2 - 1\right) = 4(x - 1)^2 - 4\]

### Step 2: Completing the square for \(y\)

For terms involving \(y\):
\[y^2 - 6y = (y - 3)^2 - 9\]
Thus:
\[y^2 - 6y = (y - 3)^2 - 9\]

### Step 3: Substitute back into the PDF

Substituting back the completed squares:
\[
f_{X,Y}(x,y) = c \cdot \exp\left(-\frac{1}{2} \left(4(x - 1)^2 - 4 + (y - 3)^2 - 9 + 13\right)\right)
\]
Simplifying further:
\[
= c \cdot \exp\left(-\frac{1}{2} \left(4(x - 1)^2 + (y - 3)^2 + 0\right)\right)
= c \cdot \exp\left(-2(x - 1)^2 - \frac{1}{2}(y - 3)^2\right)
\]

### Step 4: Recognizing the joint distribution

This joint PDF represents a bivariate normal distribution:
\[
f_{X,Y}(x,y) = c \cdot \exp\left(-\frac{1}{2} \left( z^T A z \right)\right)
\]
where:
- The \(z\) vector contains the deviations from the means of the individual distributions.
- \(A\) specifies the variances and covariance.

From the form \(f_{X,Y}(x,y)\):
- The marginal distribution of \(X\) is \(N(1, \frac{1}{4})\) (mean \(1\), variance \(\frac{1}{4}\) since \(2=\frac{1}{\sigma_x^2}\) leads to \(\sigma_x^2 = \frac{1}{4}\)).
- The marginal distribution of \(Y\) is \(N(3, 2)\) (mean \(3\), variance \(2\) since \(\frac{1}{2} = \frac{1}{\sigma_y^2}\) leads to \(\sigma_y^2 = 2\)).

### Step 5: Calculate expected values and variances

Using properties of the Gaussian distribution:

1. **Expected value \(E[X]\)**:
\[
E[X] = 1
\]

2. **Variance \(Var(X)\)**:
\[
Var(X) = \frac{1}{4}
\]

3. **Expected value \(E[Y]\)**:
\[
E[Y] = 3
\]

4. **Variance \(Var(Y)\)**:
\[
Var(Y) = 2
\]

### Final Results

- \(E[X] = 1\)
- \(Var(X) = \frac{1}{4}\)
- \(E[Y] = 3\)
- \(Var(Y) = 2\)