1. **Problem:** Show that the expected value of a constant $k$ is $k$. Step 1: By definition, the expected value $E(k)$ is the sum over all outcomes of $k$ times their probabilities. Step 2: Since $k$ is constant, $E(k) = \sum k \cdot P(\text{outcome}) = k \sum P(\text{outcome})$. Step 3: The sum of all probabilities is 1, so $E(k) = k \times 1 = k$. Example: If you always get 10, $E(10) = 10$. 2. **Problem:** Show that the variance of a constant $k$ is 0. Step 1: Variance is defined as $\mathrm{Var}(k) = E[(k - E(k))^2]$. Step 2: Since $E(k) = k$, this becomes $E[(k - k)^2] = E[0^2] = 0$. Example: If your income is always 50, variance is 0. 3. **Problem:** Show that $E(kX) = k E(X)$ for a constant $k$ and random variable $X$. Step 1: By definition, $E(kX) = \sum (k x_i) P(x_i)$. Step 2: Factor out $k$: $E(kX) = k \sum x_i P(x_i) = k E(X)$. Example: For a die roll $X$ with $E(X) = 3.5$, if $k=2$, then $E(2X) = 2 \times 3.5 = 7$. 4. **Problem:** Show that $\mathrm{Var}(kX) = k^2 \mathrm{Var}(X)$. Step 1: By definition, $\mathrm{Var}(kX) = E[(kX - kE(X))^2]$. Step 2: Factor out $k^2$: $E[k^2 (X - E(X))^2] = k^2 E[(X - E(X))^2] = k^2 \mathrm{Var}(X)$. Example: For die roll variance $\approx 2.92$, if $k=2$, then variance is $4 \times 2.92 = 11.68$. 5. **Problem:** Show that $E(X + Y) = E(X) + E(Y)$. Step 1: By definition, $E(X + Y) = \sum (x_i + y_j) P(x_i, y_j)$. Step 2: Distribute sum: $= \sum x_i P(x_i, y_j) + \sum y_j P(x_i, y_j)$. Step 3: These are marginal expectations, so $E(X + Y) = E(X) + E(Y)$. Example: For two coin flips with $E(X) = 0.5$ and $E(Y) = 0.5$, total expected heads is 1. Final answers: $E(k) = k$ $\mathrm{Var}(k) = 0$ $E(kX) = k E(X)$ $\mathrm{Var}(kX) = k^2 \mathrm{Var}(X)$ $E(X + Y) = E(X) + E(Y)$