Answer

Part (a)

First find the marginal totals:
$P(X=1) = 0.2 + 0.3 = 0.5$
$P(Y=1) = 0.2 + 0.1 = 0.3$
The product of their marginals is: $P(X=1) \times P(Y=1) = 0.5 \times 0.3 = 0.15$.
Looking at the table, the joint probability is $P(X=1, Y=1) = 0.2$. Since $0.2 \neq 0.15$, $X$ and $Y$ are not independent.

Part (b)

Summing across rows:
$P(X = 1) = 0.2 + 0.3 = 0.5$
$P(X = 2) = 0.1 + 0.4 = 0.5$

Part (c)

Divide each joint probability in row $X=1$ by $P(X=1) = 0.5$:
$P(Y = 1 \mid X = 1) = \frac{0.2}{0.5} = 0.4$
$P(Y = 2 \mid X = 1) = \frac{0.3}{0.5} = 0.6$

Part (d)

$E[Y \mid X = 1] = (1 \times 0.4) + (2 \times 0.6) = 0.4 + 1.2 = 1.6$

Answer

Part (a)

First find the marginal totals:
• Row sums: $P(X=5) = 0.20$, $P(X=10) = 0.80$
• Column sums: $P(Y=0) = 0.70$, $P(Y=4) = 0.30$
Checking the elements:
$P(X=5)P(Y=0) = 0.20 \times 0.70 = 0.14 = P(X=5, Y=0)$
$P(X=10)P(Y=4) = 0.80 \times 0.30 = 0.24 = P(X=10, Y=4)$
Since $P(X=x, Y=y) = P(X=x)P(Y=y)$ holds true for every cell, $X$ and $Y$ are independent.

Part (b)

• $P(X = 5) = 0.2$
• $P(X = 10) = 0.8$

Part (c)

Since $X$ and $Y$ are independent, it matches its marginal distribution:
• $P(Y = 0 \mid X = 10) = 0.70$
• $P(Y = 4 \mid X = 10) = 0.30$

Part (d)

$E[Y \mid X = 10] = (0 \times 0.70) + (4 \times 0.30) = 0 + 1.2 = 1.2$

Answer

Part (a)

Notice that $P(X=2, Y=1) = 0$. However, $P(X=2) = 0.50$ and $P(Y=1) = 0.25$.
Since $P(X=2)P(Y=1) = 0.125 \neq 0$, $X$ and $Y$ are not independent.

Part (b)

• $P(X = 0) = 0.25 + 0.25 = 0.50$
• $P(X = 2) = 0.50 + 0.00 = 0.50$

Part (c)

Divide row values by $P(X=0) = 0.50$:
• $P(Y = -1 \mid X = 0) = \frac{0.25}{0.50} = 0.50$
• $P(Y = 1 \mid X = 0) = \frac{0.25}{0.50} = 0.50$

Part (d)

$E[Y \mid X = 0] = (-1 \times 0.50) + (1 \times 0.50) = -0.50 + 0.50 = 0.0$

Answer

Part (a)

To check if $X$ and $Y$ are independent, we look at the marginal probabilities.
$P(X=1) = 0.1 + 0.2 + 0.1 = 0.4$
$P(Y=3) = 0.2 + 0.0 + 0.1 = 0.3$
The product of their marginals is: $P(X=1) \times P(Y=3) = 0.4 \times 0.3 = 0.12$.
Looking at the table, the joint probability is $P(X=1, Y=3) = 0.2$. Since $0.2 \neq 0.12$, $X$ and $Y$ are not independent.

Part (b)

Summing across the rows for each value of $X$:
• $P(X = 1) = 0.1 + 0.2 + 0.1 = 0.4$
• $P(X = 2) = 0.3 + 0.0 + 0.1 = 0.4$
• $P(X = 3) = 0.1 + 0.1 + 0.0 = 0.2$

Part (c)

We divide each joint probability in row $X=1$ by $P(X=1) = 0.4$:
• $P(Y = 1 \mid X = 1) = \frac{0.1}{0.4} = 0.25$
• $P(Y = 3 \mid X = 1) = \frac{0.2}{0.4} = 0.50$
• $P(Y = 5 \mid X = 1) = \frac{0.1}{0.4} = 0.25$

Part (d)

$E[Y \mid X = 1] = (1 \times 0.25) + (3 \times 0.50) + (5 \times 0.25) = 0.25 + 1.50 + 1.25 = 3.0$

Answer

Part (a)

Let's find the marginal totals:
• Row sums: $P(X=0) = 0.40$, $P(X=5) = 0.60$
• Column sums: $P(Y=0) = 0.30$, $P(Y=2) = 0.45$, $P(Y=4) = 0.25$
Checking the cells:
$P(X=0)P(Y=0) = 0.40 \times 0.30 = 0.12 = P(X=0, Y=0)$
$P(X=5)P(Y=4) = 0.60 \times 0.25 = 0.15 = P(X=5, Y=4)$
Since $P(X=x, Y=y) = P(X=x)P(Y=y)$ holds true for every cell, $X$ and $Y$ are independent.

Part (b)

• $P(X = 0) = 0.12 + 0.18 + 0.10 = 0.40$
• $P(X = 5) = 0.18 + 0.27 + 0.15 = 0.60$

Part (c)

Since $X$ and $Y$ are independent, the conditional distribution of $Y$ given $X=5$ is identical to its marginal distribution:
• $P(Y = 0 \mid X = 5) = P(Y = 0) = 0.30$
• $P(Y = 2 \mid X = 5) = P(Y = 2) = 0.45$
• $P(Y = 4 \mid X = 5) = P(Y = 4) = 0.25$

Part (d)

$E[Y \mid X = 5] = (0 \times 0.30) + (2 \times 0.45) + (4 \times 0.25) = 0 + 0.90 + 1.00 = 1.9$

Answer

Part (a)

Let's find the marginal totals:
• Row sums: $P(X=2) = 0.25$, $P(X=4) = 0.50$, $P(X=6) = 0.25$
• Column sums: $P(Y=10) = 0.20$, $P(Y=20) = 0.60$, $P(Y=30) = 0.20$
Testing the cells:
$P(X=2)P(Y=10) = 0.25 \times 0.20 = 0.05 = P(X=2, Y=10)$
$P(X=4)P(Y=20) = 0.50 \times 0.60 = 0.30 = P(X=4, Y=20)$
Every single joint cell satisfies the product condition, so $X$ and $Y$ are independent.

Part (b)

• $P(X = 2) = 0.25$
• $P(X = 4) = 0.50$
• $P(X = 6) = 0.25$

Part (c)

Due to independence, it matches the marginal distribution of $Y$:
• $P(Y = 10 \mid X = 4) = 0.20$
• $P(Y = 20 \mid X = 4) = 0.60$
• $P(Y = 30 \mid X = 4) = 0.20$

Part (d)

$E[Y \mid X = 4] = (10 \times 0.20) + (20 \times 0.60) + (30 \times 0.20) = 2 + 12 + 6 = 20.0$

Answer

Part (a)

Using the linearity of expectation:
$E[3X - Y] = 3E[X] - E[Y] = 3(3) - 5 = 9 - 5 = 4$

Part (b)

Because $X$ and $Y$ are independent, the covariance term is zero. Note that the coefficient of $Y$ is $-1$, and $(-1)^2 = 1$:
$Var(3X - Y) = 3^2 Var(X) + (-1)^2 Var(Y) = 9(4) + 1(2) = 36 + 2 = 38$

Answer

Part (a)

Using the linearity of expectation:
$E[X + 4Y] = E[X] + 4E[Y] = 10 + 4(4) = 10 + 16 = 26$

Part (b)

Since $X$ and $Y$ are independent, the variance of the linear combination is:
$Var(X + 4Y) = Var(X) + 4^2 Var(Y) = 9 + 16(16) = 9 + 256 = 265$

Answer

Part (a)

Extending linearity of expectation to three variables:
$E[X + 2Y - 3Z] = E[X] + 2E[Y] - 3E[Z] = 2 + 2(6) - 3(1) = 2 + 12 - 3 = 11$

Part (b)

Since the variables are mutually independent, we square the coefficients (including the negative coefficient, where $(-3)^2 = 9$):
$Var(X + 2Y - 3Z) = Var(X) + 2^2 Var(Y) + (-3)^2 Var(Z)$
$Var(X + 2Y - 3Z) = 3 + 4(5) + 9(8) = 3 + 20 + 72 = 95$

Answer

Part (a)

$\rho = \frac{-1000}{\sqrt{1600} \times \sqrt{2500}} = \frac{-1000}{40 \times 50} = \frac{-1000}{2000} = -0.5$

Part (b)

$E[X + Y] = 50 + 50 = 100$

Part (c)

$Var(X + Y) = 1600 + 2500 + 2(-1000) = 4100 - 2000 = 2100$

Answer

Part (a)

The correlation is the covariance divided by the standard deviations, i.e. $-120 / (10 \times 20) = -120 / 200 = -0.6$.

Part (b)

$EX = EY = 20$, so $E[X + Y] = EX + EY = 20 + 20 = 40$.

Part (c)

The variance of $(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y) = 100 + 400 - 240 = 260$.