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Let's break down the problem step by step.

### Part 1: Valid Probability Mass Function
To determine the values of \( d \) for which \( p_X(x) \) describes a valid probability mass function (PMF), we need to satisfy two key conditions:
1. **Non-negativity:** Each probability value must be non-negative, i.e., \( p_X(x) \geq 0 \) for all \( x \).
2. **Normalization:** The sum of all the probabilities must equal 1, i.e., \( \sum_{x=1}^{4} p_X(x) = 1 \).

The given probabilities are:
\[
p_X(1) = \frac{1 + 3d}{4}, \quad p_X(2) = \frac{1 - d}{4}, \quad p_X(3) = \frac{1 + 2d}{4}, \quad p_X(4) = \frac{1 - 4d}{4}
\]

#### Step 1: Non-negativity
Each of these expressions must be greater than or equal to 0:

\[
\frac{1 + 3d}{4} \geq 0 \quad \Rightarrow \quad 1 + 3d \geq 0 \quad \Rightarrow \quad d \geq -\frac{1}{3}
\]

\[
\frac{1 - d}{4} \geq 0 \quad \Rightarrow \quad 1 - d \geq 0 \quad \Rightarrow \quad d \leq 1
\]

\[
\frac{1 + 2d}{4} \geq 0 \quad \Rightarrow \quad 1 + 2d \geq 0 \quad \Rightarrow \quad d \geq -\frac{1}{2}
\]

\[
\frac{1 - 4d}{4} \geq 0 \quad \Rightarrow \quad 1 - 4d \geq 0 \quad \Rightarrow \quad d \leq \frac{1}{4}
\]

Combining these, we get the range for \( d \):
\[
-\frac{1}{3} \leq d \leq \frac{1}{4}
\]

#### Step 2: Normalization
Now, we sum all the probabilities and set it equal to 1:
\[
\frac{1 + 3d}{4} + \frac{1 - d}{4} + \frac{1 + 2d}{4} + \frac{1 - 4d}{4} = 1
\]
Simplifying the left-hand side:
\[
\frac{(1 + 3d) + (1 - d) + (1 + 2d) + (1 - 4d)}{4} = 1
\]
\[
\frac{4 + 0d}{4} = 1
\]
\[
1 = 1
\]
Thus, the normalization condition is satisfied for all \( d \), and the only constraint on \( d \) comes from the non-negativity condition.

Hence, the valid range for \( d \) is:
\[
-\frac{1}{3} \leq d \leq \frac{1}{4}
\]

### Part 2: Minimizing Variance
The variance of \( X \), \( \text{Var}(X) \), is given by:
\[
\text{Var}(X) = E[X^2] - (E[X])^2
\]
where \( E[X] \) is the expected value and \( E[X^2] \) is the second moment.

#### Step 1: Compute \( E[X] \)
The expected value \( E[X] \) is:
\[
E[X] = 1 \cdot p_X(1) + 2 \cdot p_X(2) + 3 \cdot p_X(3) + 4 \cdot p_X(4)
\]
Substitute the given probabilities:
\[
E[X] = 1 \cdot \frac{1 + 3d}{4} + 2 \cdot \frac{1 - d}{4} + 3 \cdot \frac{1 + 2d}{4} + 4 \cdot \frac{1 - 4d}{4}
\]
Simplifying:
\[
E[X] = \frac{1 + 3d}{4} + \frac{2(1 - d)}{4} + \frac{3(1 + 2d)}{4} + \frac{4(1 - 4d)}{4}
\]
\[
E[X] = \frac{(1 + 3d) + 2(1 - d) + 3(1 + 2d) + 4(1 - 4d)}{4}
\]
Expanding and combining like terms:
\[
E[X] = \frac{1 + 3d + 2 - 2d + 3 + 6d + 4 - 16d}{4}
\]
\[
E[X] = \frac{10 - 9d}{4}
\]

#### Step 2: Compute \( E[X^2] \)
Similarly, we compute \( E[X^2] \):
\[
E[X^2] = 1^2 \cdot p_X(1) + 2^2 \cdot p_X(2) + 3^2 \cdot p_X(3) + 4^2 \cdot p_X(4)
\]
\[
E[X^2] = 1 \cdot \frac{1 + 3d}{4} + 4 \cdot \frac{1 - d}{4} + 9 \cdot \frac{1 + 2d}{4} + 16 \cdot \frac{1 - 4d}{4}
\]
Simplifying:
\[
E[X^2] = \frac{1 + 3d + 4(1 - d) + 9(1 + 2d) + 16(1 - 4d)}{4}
\]
\[
E[X^2] = \frac{1 + 3d + 4 - 4d + 9 + 18d + 16 - 64d}{4}
\]
\[
E[X^2] = \frac{30 - 47d}{4}
\]

#### Step 3: Variance
Now, calculate the variance:
\[
\text{Var}(X) = E[X^2] - (E[X])^2
\]
Substitute the values for \( E[X^2] \) and \( E[X] \):
\[
\text{Var}(X) = \frac{30 - 47d}{4} - \left( \frac{10 - 9d}{4} \right)^2
\]
Simplifying:
\[
\text{Var}(X) = \frac{30 - 47d}{4} - \frac{(10 - 9d)^2}{16}
\]
\[
\text{Var}(X) = \frac{30 - 47d}{4} - \frac{100 - 180d + 81d^2}{16}
\]
\[
\text{Var}(X) = \frac{4(30 - 47d) - (100 - 180d + 81d^2)}{16}
\]
\[
\text{Var}(X) = \frac{120 - 188d - 100 + 180d - 81d^2}{16}
\]
\[
\text{Var}(X) = \frac{20 - 8d - 81d^2}{16}
\]
\[
\text{Var}(X) = \frac{-81d^2 - 8d + 20}{16}
\]

#### Step 4: Minimize the Variance
To minimize the variance, we take the derivative of \( \text{Var}(X) \) with respect to \( d \) and set it equal to 0:
\[
\frac{d}{dd} \left( \frac{-81d^2 - 8d + 20}{16} \right) = 0
\]
\[
\frac{-162d - 8}{16} = 0
\]
\[
-162d - 8 = 0
\]
\[
d = -\frac{8}{162} = -\frac{4}{81}
\]

Thus, the value of \( d \) that minimizes the variance is \( d = -\frac{4}{81} \).

### Final Answer:
- The valid range for \( d \) is \( -\frac{1}{3} \leq d \leq \frac{1}{4} \).
- The value of \( d \) that minimizes the variance is \( d = -\frac{4}{81} \).

---

Would you like a further breakdown of any part of this solution?

Here are some related questions to explore:
1. How do we verify the normalization condition for a probability mass function?
2. What is the significance of minimizing variance in probability distributions?
3. Can variance be negative? Why or why not?
4. How does changing the value of \( d \) affect the probabilities and distribution of \( X \)?
5. What is the intuition behind expected value and variance in

Let X be a discrete random variable with p_X(1) = (1 + 3d)/4, p_X(2) = (1 - d)/4, p_X(3) = (1 + 2d)/4, and p_X(4) = (1 - 4d)/4. For what values of d does p_X(x) describe a valid probability mass function? Further determine the value of d for which Var(X) is a minimum.

Solve for the valid values of d for which the given probability mass function p_X(x) is non-negative and normalized. Further, calculate the value of d that minimizes the variance Var(X) using the provided probabilities. Includes detailed steps in probability and variance calculations.

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