Determine Whether Or Not The Function Is A Probability Density Function Over The Given Interval. 3 F(x)

Determine Whether Or Not The Function Is A Probability Density Function Over The Given Interval. 3 F(x)

When analyzing whether a function qualifies as a probability density function (PDF), it is essential to verify specific criteria that define the properties of PDFs in probability theory. The function in question, denoted as 3F(x), suggests a scaled version of some original function F(x). To ascertain if 3F(x) is a valid PDF over a given interval, one must examine its properties carefully. This involves ensuring that the function is non-negative over its domain and that the total integral over the specified interval equals 1. This article will guide you through the process of determining whether 3F(x) meets these criteria, discussing the necessary steps, mathematical conditions, and implications involved in the verification process.

Understanding Probability Density Functions (PDFs)

Definition and Properties of a PDF

A probability density function (PDF) is a fundamental concept in probability and statistics used to describe the likelihood of a continuous random variable taking on a particular value within a given range. Unlike discrete probability distributions, which assign probabilities to specific points, PDFs define the density of probability across a continuum.

The key properties of a PDF, f(x), are:

• Non-negativity: f(x) ≥ 0 for all x in the domain.
• Normalization: The total area under the curve of f(x) over its domain is 1, i.e.,
  ∫_a^b f(x) dx = 1, where [a, b] is the interval of support.

Any function satisfying these properties can be considered a valid PDF. The second property ensures that the total probability across the entire domain sums to 1, aligning with the fundamental axioms of probability.

Implications of Scaling a PDF by a Constant

Scaling a PDF by a constant factor alters its total area. For example, multiplying a function F(x) by 3 results in a new function 3F(x). Whether this new function remains a valid PDF depends on whether the scaled area still sums to 1 over the specified interval.

If F(x) is a PDF, then:

• ∫_a^b F(x) dx = 1


• The scaled function, 3F(x), will have an integral:

∫_a^b 3F(x) dx = 3 ∫_a^b F(x) dx = 3 × 1 = 3

which violates the normalization condition for a PDF. Therefore, unless the scaling factor is 1, the resulting function generally does not qualify as a PDF unless additional normalization is performed.

Step-by-Step Procedure to Verify Whether 3F(x) is a PDF

1. Identify the Support Interval

Before performing any calculations, determine the interval over which the function is defined. This is crucial because the properties of the function are only relevant within its support.

• Check the problem statement or given data for the domain of F(x).


• Note any restrictions or discontinuities that may influence the integral.

2. Verify Non-Negativity of 3F(x)

Since a PDF must be non-negative over its support:

• Check the original function F(x): Is F(x) ≥ 0 for all x in the interval?


• Determine the sign of 3F(x): Multiplying by 3 does not change the sign; if F(x) ≥ 0, then 3F(x) ≥ 0.

If F(x) takes negative values at any point in the interval, then 3F(x) will also be negative there, disqualifying it as a PDF.

3. Calculate the Integral of 3F(x) Over the Support

The core criterion involves the total integral:

• Compute ∫_a^b 3F(x) dx


• Use the properties of integrals:

∫_a^b 3F(x) dx = 3 ∫_a^b F(x) dx

• Evaluate this integral either analytically or numerically based on the form of F(x).

4. Check for Normalization

• If ∫_a^b 3F(x) dx = 1, then 3F(x) satisfies the normalization condition.


• If the integral is greater than or less than 1, then 3F(x) does not qualify as a PDF in its current form.

Note: Since the integral of F(x) over the support is generally fixed (for example, if F(x) is a PDF), multiplying by 3 typically results in the integral being 3, which exceeds 1. Therefore, unless F(x) has an integral of 1/3 over its support, 3F(x) will not be a valid PDF without normalization.

Interpreting the Results and Additional Considerations

Case 1: F(x) is a Valid PDF and the integral of F(x) over its support is 1

• The integral of 3F(x) over the same support is 3.
• Therefore, 3F(x) is not a valid PDF because it does not satisfy the normalization condition.
• Solution: Normalize 3F(x) by dividing by 3:

g(x) = (3F(x)) / 3 = F(x) (which is the original PDF).

• Alternatively, consider scaling F(x) with a factor such that the total area is 1.

Case 2: F(x) is not a PDF, or its integral over the support is not 1

• The integral of 3F(x) might be less than 1, equal to 1, or greater than 1.
• If the integral is less than 1: 3F(x) is not a PDF unless scaled appropriately.
• If the integral is greater than 1: same applies; normalization is necessary.

Implications of Scaling and Normalization

• The act of multiplying by 3 is a simple scaling, but for a function to qualify as a PDF, the total area must be exactly 1.
• If you desire to create a valid PDF from 3F(x), you must normalize by dividing by the total integral:

f_normalized(x) = (3F(x)) / ∫_a^b 3F(x) dx

• This normalization ensures the total area under the curve is 1.

Practical Examples and Applications

Example 1: F(x) = (1/2) over [0, 2]

Suppose F(x) = (1/2) for x in [0, 2], zero elsewhere.

• F(x) is a valid PDF because:


• Non-negative over [0, 2].


• Total area: ∫₀² (1/2) dx = (1/2) × 2 = 1.


• Now, consider 3F(x):


• Integral over [0, 2]:

∫₀² 3F(x) dx = 3 × (1/2) × 2 = 3 × 1 = 3.

• Since the total area is 3, 3F(x) is not a PDF unless scaled down.


• To normalize:

g(x) = (3F(x)) / 3 = F(x).

• Thus, the original F(x) is a valid PDF, but 3F(x) is not unless further normalization is applied.

Application: This example illustrates why scaling a PDF by a constant other than 1 generally invalidates it as a PDF unless normalized.

Example 2: F(x) = x/3 over [0, 3]

• Calculate integral:

∫₀³ (x/3) dx = (1/3) × (x²/2) |₀³ = (1/3) × (9/2) = (1/3) × 4.5 = 1.5.

• Since the integral is 1.5, F(x) is not a valid PDF, but scaled appropriately:

F(x) / 1.5 would be a valid PDF over [0, 3].

• Now, consider 3F(x):

∫₀³ 3F