Problem 89 - Applied Statistics and Probability for Engineers

Question

The computational time of a statistical analysis applied to a data set can sometimes increase with the square of \(N,\) the number of rows of data. Suppose that for a particular algorithm, the computation time is approximately \(T=0.004 N^{2}\) seconds. Although the number of rows is a discrete measurement, assume that the distribution of \(N\) over a number of data sets can be approximated with an exponential distribution with a mean of 10,000 rows. Determine the probability density function and the mean of \(T\).

Step-by-Step Solution

Answer

The PDF of \(T\) needs further simplification; the mean of \(T\) is 440,000 seconds.

Step 1: Understand the Problem

We need to determine the probability density function (PDF) of the computation time, \(T = 0.004N^2\), where \(N\) has an exponential distribution with a mean of 10,000. We also need to find the mean of \(T\).

Step 2: Find the PDF of Exponential Distribution

The PDF of an exponential distribution with mean \(\mu\) is given by \(f_N(n) = \frac{1}{\mu}e^{-n/\mu}\). Here, \(\mu = 10,000\), so \(f_N(n) = \frac{1}{10,000}e^{-n/10,000}\) for \(n \geq 0\).

Step 3: Transform the Variable

The transformation of \(N\) to \(T\) is given by \(T = 0.004N^2\). To find the PDF of \(T\), we utilize the method of transformation of variables. Since the transformation \(T = g(N) = 0.004N^2\) is non-linear, we'll use the change of variables technique.

Step 4: Find the PDF of T

The transformation is \(T = 0.004N^2\), so \(N = \sqrt{\frac{T}{0.004}}\). The derivative of this transformation with respect to \(T\) is \(\frac{dN}{dT} = \frac{1}{2\sqrt{0.004T}}\). Substitute \(N = \sqrt{\frac{T}{0.004}}\) into \(f_N(n)\) and multiply by \(\left|\frac{dN}{dT}\right|\): \[f_T(t) = f_N\left(\sqrt{\frac{t}{0.004}}\right) \cdot \left|\frac{dN}{dT}\right| = \frac{\sqrt{0.004}}{10,000} e^{\frac{-\sqrt{t/0.004}}{10,000}} \, \frac{1}{2\sqrt{0.004t}}\]

Step 5: Calculate the Mean of T

The mean of \(T\), denoted by \(E[T]\), can be found using the transformation of expectation, knowing that \(E[T] = E[0.004N^2]\). Since the mean of \(N\) is 10,000, and the mean of a function of \(N\) is \(E[T] = 0.004 \times E[N^2]\) involves the second moment of the exponential distribution. For an exponential distribution, \(E[N^2] = \mu(\mu + 1) = 10,000 \times 11,000\). Thus, \[E[T] = 0.004 \times 10,000 \times 11,000\] Finally, \[E[T] = 440,000\] seconds.

Key Concepts

Probability Density Function

The Probability Density Function (PDF) is a fundamental concept in statistics that describes the likelihood of a random variable taking on a specific value.
For continuous random variables, the PDF provides a way to calculate probabilities over a range of values but not the probability of a precise value.
In the problem, the number of rows of data, denoted as \(N\), follows an exponential distribution. This means the PDF of \(N\) is given by:

• \( f_N(n) = \frac{1}{\mu} e^{-n/\mu} \) where \( \mu \) is the mean.

Here, \( \mu = 10,000 \), indicating how the data rows are distributed in this statistical analysis.
The exponential distribution is often applied to model times until an event occurs, like computation duration in our example.
This PDF provides a valuable function because it relates how changes in the number of rows affect computation time.
Knowing the PDF allows us to predict and understand how the variable behaves within its range.

Transformation of Variables

The Transformation of Variables is a technique used to find the PDF of a derived random variable from another variable.
This method helps when you have a non-linear function of a variable, like the computation time \(T = 0.004N^2\).
To find the PDF of \(T\), we use the relationship between \(T\) and \(N\). Here are the steps simplified:

• We know \(T = 0.004N^2\), which is a non-linear transformation.
• Solve for \(N\) in terms of \(T\): \(N = \sqrt{\frac{T}{0.004}}\).
• Calculate the derivative: \(\frac{dN}{dT} = \frac{1}{2\sqrt{0.004T}}\).
• The PDF of \(T\), \(f_T(t)\), is calculated by substituting \(N\) with the derived expression and multiplying by the absolute value of the derivative.

This transformation helps represent how the distribution of data rows affects the distribution of computation time.
Understanding this concept is crucial as it allows us to handle situations where variables are interconnected in complex, non-linear relationships.

Expectation and Mean Calculation

Expectation and mean are central concepts in statistics, describing the average outcome of a random variable.
In our scenario, the goal was to find the mean computational time \(E[T]\), given as \(T = 0.004N^2\).
Firstly, recall the mean of an exponential distribution for \(N\) is \(10,000\).
To find \(E[T]\), our transformation gives:

• \(E[T] = E[0.004N^2]\).
• The second moment for an exponentially distributed variable is \(E[N^2] = \mu(\mu + 1)\).
• Substitute: \(E[N^2] = 10,000 \times 11,000\).
• The final calculation provides \(E[T] = 0.004 \times 10,000 \times 11,000 = 440,000\).

This calculation demonstrates how transformations impact expectations.
By understanding the expected value, you can predict average outcomes and make informed decisions based on the statistical nature of the variable.