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Chapter 4: Problem 12

Let \(f_{1}(x)\) and \(f_{2}(x)\) be probability mass functions. Show that if \(0 \leq p \leq 1\), then \(f_{3}(x)=p f_{1}(x)+\) \((1-p) f_{2}(x)\) is a probability mass function. Interpret this result.

Short Answer

Expert verified
\(f_3(x)\) is a PMF as it satisfies non-negativity and normalization. It's a mixture of \(f_1(x)\) and \(f_2(x)\).

Step by step solution

01

Verify Non-negativity

Any probability mass function (PMF) must satisfy the non-negativity condition, meaning \(f_1(x) \geq 0\) and \(f_2(x) \geq 0\) for all \(x\). Since \(0 \leq p \leq 1\), it follows that \(p f_1(x) \geq 0\) and \((1-p) f_2(x) \geq 0\). Therefore, \(f_3(x) = p f_1(x) + (1-p) f_2(x) \geq 0\) for all \(x\).
02

Verify Normalization Condition

The sum of a PMF over all possible values of \(x\) should be equal to 1, i.e., \(\sum_x f_3(x) = 1\). Start by expanding \(f_3(x)\): \[ \sum_x f_3(x) = \sum_x (p f_1(x) + (1-p) f_2(x)) = p \sum_x f_1(x) + (1-p) \sum_x f_2(x) \]Given that \(f_1(x)\) and \(f_2(x)\) are PMFs, \(\sum_x f_1(x) = 1\) and \(\sum_x f_2(x) = 1\); thus, \[ p \cdot 1 + (1-p) \cdot 1 = p + (1-p) = 1 \]. Thus, \(f_3(x)\) is normalized.
03

Interpretation

The function \(f_3(x) = p f_1(x) + (1-p) f_2(x)\) is a weighted average of the two PMFs \(f_1(x)\) and \(f_2(x)\). The parameter \(p\) determines the contribution of each PMF to the mixture: when \(p = 1\), \(f_3(x) = f_1(x)\); when \(p = 0\), \(f_3(x) = f_2(x)\). Hence, \(f_3(x)\) is a convex combination of two valid PMFs and remains a PMF.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-Negativity Condition
The non-negativity condition is a fundamental property that every probability mass function (PMF) must satisfy. This condition ensures that for any value of a random variable, the probability is always non-negative. It is stated as follows:
  • For any PMF \(f(x)\), \(f(x) \geq 0\) for all \(x\).
When discussing a mixture of PMFs, such as \(f_3(x) = p f_1(x) + (1-p) f_2(x)\), the non-negativity condition must hold true for this new function as well.
In the original exercise, both \(f_1(x)\) and \(f_2(x)\) are assumed to be PMFs. Therefore, each function satisfies the condition \(f_1(x) \geq 0\) and \(f_2(x) \geq 0\) for all \(x\). Additionally, since \(0 \leq p \leq 1\), each term \(p f_1(x)\) and \((1-p) f_2(x)\) will also be non-negative. Adding two non-negative quantities \(p f_1(x) + (1-p) f_2(x)\) will also result in a non-negative value for \(f_3(x)\). Thus, \(f_3(x)\) meets the non-negativity condition.
Normalization Condition
The normalization condition is another essential property of a PMF, ensuring that the total probability across all possible outcomes of a random variable sums to one. Mathematically, this condition is expressed as \(\sum_x f(x) = 1\) for a PMF \(f(x)\).
In the case of \(f_3(x) = p f_1(x) + (1-p) f_2(x)\), checking normalization involves verifying that the entire sum over all \(x\) values remains 1.
Given:
  • \(\sum_x f_1(x) = 1\)
  • \(\sum_x f_2(x) = 1\)
We expand and check: \[ \sum_x f_3(x) = \sum_x (p f_1(x) + (1-p) f_2(x))\] Now, split the sums: \[ p \sum_x f_1(x) + (1-p) \sum_x f_2(x) = p \cdot 1 + (1-p) \cdot 1 = p + (1-p) = 1 \]
Thus, \(f_3(x)\) is normalized, confirming it behaves as a valid PMF with the sum of probabilities equal to one.
Convex Combination
A convex combination is a linear combination of points (or in this case, functions) where the coefficients (or weights) are non-negative and sum up to one. This concept is important when mixing or blending different functions or distributions to create another one that retains some properties of the originals.
In the provided PMF mixture \(f_3(x) = p f_1(x) + (1-p) f_2(x)\), the value \(p\) serves as a weight that determines how much influence each function \(f_1(x)\) and \(f_2(x)\) has on the mixture. The conditions \(0 \leq p \leq 1\) ensure that this mixture is a convex combination:
  • \(p\) is non-negative and weighs the contribution of \(f_1(x)\).
  • \(1-p\) is non-negative and weighs the contribution of \(f_2(x)\).
  • The sum of \(p + (1-p) = 1\) keeps the mixture's overall probability intact.
As a result, the blended function \(f_3(x)\) remains within the constraints of a PMF, accrediting the definition as a probability mass function itself. This kind of combination allows for flexibility in modeling scenarios where outcomes can be represented as a blend of existing probabilistic models, providing a nuanced distribution that covers more real-world situations.

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