To solve the given problem, we will calculate each part step-by-step:

Part (a) $E[X]$

The expected value of $X$ is calculated as: $E[X] = \sum_{x} x \cdot P(X = x)$ To compute $P(X = x)$, sum over the joint probabilities for all $Y$ values:

• $P(X=0) = 0.107 + 0.214 + 0.037 = 0.358$
• $P(X=1) = 0.321 + 0.214 + 0.000 = 0.535$
• $P(X=2) = 0.107 + 0.000 + 0.000 = 0.107$

Now compute $E[X]$: $E[X] = (0 \cdot 0.358) + (1 \cdot 0.535) + (2 \cdot 0.107) = 0 + 0.535 + 0.214 = 0.749$

Part (b) $\text{Var}[X]$

Variance is calculated as: [ \text{Var}[X] = E[X^2] - (E[X])^2 ] First, compute $E[X^2]$: $E[X^2] = \sum_{x} x^2 \cdot P(X=x) = (0^2 \cdot 0.358) + (1^2 \cdot 0.535) + (2^2 \cdot 0.107)$ $E[X^2] = 0 + 0.535 + 4 \cdot 0.107 = 0.535 + 0.428 = 0.963$ Now, substitute $E[X] = 0.749$: $\text{Var}[X] = 0.963 - (0.749)^2 = 0.963 - 0.561 = 0.402$

Part (c) $E[Y \mid X=1]$

The conditional expectation is: $E[Y \mid X=1] = \sum_{y} y \cdot P(Y = y \mid X=1)$ First, compute $P(Y = y \mid X=1)$:

• $P(Y=0 \mid X=1) = \frac{P(X=1, Y=0)}{P(X=1)} = \frac{0.321}{0.535} = 0.6$
• $P(Y=1 \mid X=1) = \frac{P(X=1, Y=1)}{P(X=1)} = \frac{0.214}{0.535} = 0.4$
• $P(Y=2 \mid X=1) = \frac{P(X=1, Y=2)}{P(X=1)} = \frac{0.0}{0.535} = 0.0$

Now compute $E[Y \mid X=1]$: $E[Y \mid X=1] = (0 \cdot 0.6) + (1 \cdot 0.4) + (2 \cdot 0.0) = 0 + 0.4 + 0 = 0.4$

Part (d) $E[3X + 2Y]$

The linearity of expectation allows us to write: [ E[3X + 2Y] = 3E[X] + 2E[Y] ] We already know $E[X] = 0.749$. To find $E[Y]$, compute $P(Y = y)$ and use $E[Y] = \sum_{y} y \cdot P(Y = y)$:

• $P(Y=0) = 0.107 + 0.321 + 0.107 = 0.535$
• $P(Y=1) = 0.214 + 0.214 + 0.0 = 0.428$
• $P(Y=2) = 0.037 + 0.0 + 0.0 = 0.037$

Now compute $E[Y]$: $E[Y] = (0 \cdot 0.535) + (1 \cdot 0.428) + (2 \cdot 0.037) = 0 + 0.428 + 0.074 = 0.502$

Substitute into $E[3X + 2Y]$: $E[3X + 2Y] = 3(0.749) + 2(0.502) = 2.247 + 1.004 = 3.251$

Final Answers:

(a) $E[X] = 0.749$
(b) $\text{Var}[X] = 0.402$
(c) $E[Y \mid X=1] = 0.4$
(d) $E[3X + 2Y] = 3.251$

Let me know if you have any questions or want further details.

Here are 5 related questions for practice:

1. How would the calculations change if a different joint PDF were used?
2. What is the correlation coefficient between $X$ and $Y$ in this setup?
3. How do you find $P(Y \mid X)$ for all values of $X$ and $Y$?
4. What is the covariance $\text{Cov}(X, Y)$?
5. How would $E[3X + 2Y]$ change if the joint probabilities were normalized differently?

Tip: Always check the marginal probabilities $P(X)$ and $P(Y)$ to ensure they sum to 1, verifying the joint distribution is valid.