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If one picks a numerical entry at random from an almanac, or the annual accounts of a corporation, the first two significant digits, \(X\), \(Y\), are found to have approximately the joint mass function $$ f(x, y)=\log _{10}\left(1+\frac{1}{10 x+y}\right), \quad 1 \leq x \leq 9,0 \leq y \leq 9 $$ Find the mass function of \(X\) and an approximation to its mean. [A heuristic explanation for this phenomenon may be found in the second of Feller's volumes (1971).]

Short Answer

Expert verified
The mass function of \(X\) is \(f_X(x) = \sum_{y=0}^{9} \log_{10}(1 + \frac{1}{10x+y})\). The approximate mean of \(X\) needs numerical computation.

Step by step solution

01

Determine the Range of X and Y

The joint mass function describes the probability that the first two significant digits are \(X\) and \(Y\). Here, \(X\) ranges from 1 to 9, which represents the first significant digit, and \(Y\) ranges from 0 to 9, which represents the second significant digit. This means that \(X\) can take on any digit from 1 to 9, and \(Y\) can take on any digit from 0 to 9.
02

Calculate the Marginal Mass Function of X

To find the marginal mass function of \(X\), sum up the joint mass function over all possible values of \(Y\). Thus, the mass function of \(X\), denoted as \(f_X(x)\), is given by:\[ f_X(x) = \sum_{y=0}^{9} f(x, y) = \sum_{y=0}^{9} \log_{10}\left(1 + \frac{1}{10x+y}\right) \]By computing these sums for each \(x\) from 1 to 9, we can find \(f_X(x)\).
03

Compute the Approximate Mean of X

Once the mass function \(f_X(x)\) is found, compute the mean of \(X\) using the formula for the expected value:\[ E(X) = \sum_{x=1}^{9} x \cdot f_X(x) \]Evaluate this sum to find the mean of \(X\), which is an approximation to the average first significant digit.

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Key Concepts

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

Joint Mass Function
When we talk about the joint mass function in probability, it defines the probability that two random variables, such as the first two significant digits of a number, both take on specific values. In the almanac or financial accounts example, these variables are denoted by \(X\) and \(Y\). Here, \(X\) represents the first significant digit, ranging from 1 to 9, while \(Y\) is the second digit, ranging from 0 to 9.
The joint mass function \(f(x, y)\) in this case is given by \(\log_{10}\left(1 + \frac{1}{10x+y}\right)\), which means it shows how likely it is to encounter every possible combination of \(X\) and \(Y\). This framework is essential for analyzing patterns in datasets according to Benford's Law, which predicts the frequency of digits in naturally occurring datasets.
Understanding the joint mass function helps in evaluating how often each digit combination appears as the first two digits of a number, crucial for fields like data science and fraud detection.
Marginal Distribution
The marginal distribution is derived from a joint distribution. It provides the probabilities of one of the random variables without regard to the other. To achieve this, you sum the joint probabilities across possible values of the unwanted variable.
In our exercise, the marginal distribution of \(X\) involves summing the joint mass function over every possible \(Y\). Thus, the function \(f_X(x)\) is calculated by:
  • Taking each value of \(x\) from 1 to 9
  • Summing \(\log_{10}\left(1 + \frac{1}{10x+y}\right)\) for \(y\) from 0 to 9
This process simplifies the joint mass function into a collection of probabilities solely for \(X\). It tells us how likely each number is to appear as the first significant digit, independent of the second digit. Understanding the marginal distribution is useful for gaining insights into individual variables' behaviors in a dataset.
Expected Value
The expected value is a fundamental concept in probability, representing the average outcome of a random variable over many trials. It encapsulates the idea of a "long-term average."
To find the expected value of \(X\), once its marginal distribution \(f_X(x)\) is known, you multiply each value of \(x\) by its probability \(f_X(x)\) and sum the results:\[ E(X) = \sum_{x=1}^{9} x \cdot f_X(x) \]This calculation gives an average numerical value, reflecting the most probable significant digit across the chosen dataset. In the context of Benford's Law, it helps determine typical patterns and trends lurking in seemingly random numerical information.
By understanding expected value, one can predict overall outcomes and make informed decisions based on data analysis.
Numerical Data Analysis
Numerical data analysis involves examining quantitative data to uncover patterns, tendencies, and anomalies. It is a critical step in making sense of large datasets and transforming numbers into actionable insights.
One real-world application of this analysis is Benford’s Law, which helps statisticians and analysts understand the distribution of digits in natural datasets, often revealing fraud or manipulation. For instance, the unusual distribution of leading digits in accounting data could indicate anomalies or red flags indicative of irregularities.
In your dataset, this type of analysis uses the joint mass function to understand how first and second digits appear together. By studying the marginal distribution, analysts can gain a clearer picture of how often certain digits show up independently. The expected value calculation further aids in predicting or explaining the dominant trends in a dataset.
All these analyses form part of a broader toolkit required for effective data science, helping turn complex datasets into comprehensible and meaningful information.

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Most popular questions from this chapter

In your pocket is a random number \(N\) of coins, where \(N\) has the Poisson distribution with parameter \(\lambda\). You toss each coin once, with heads showing with probability \(p\) each time. Show that the total number of heads has the Poisson distribution with parameter \(\lambda p\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent random variables, and suppose that \(X_{k}\) is Bemoulli with parameter \(p_{k}\). Show that \(Y=X_{1}+X_{2}+\cdots+X_{n}\) has mean and variance given by \(\mathbf{E}(Y)=\sum_{1}^{n} p_{k}, \quad \operatorname{var}(Y)=\sum_{1}^{n} p_{k}\left(1-p_{k}\right)\) Show that, for \(\mathbb{E}(Y)\) fixed, \(\operatorname{var}(Y)\) is a maximum when \(p_{1}=p_{2}=\cdots=p_{n}\). That is to say, the variation in the sum is greatest when individuals are most alike. Is this contrary to intuition?

Let \(X\) and \(Y\) be discrete random variables with mean 0 , variance 1, and covariance \(\rho\). Show that \(\mathbb{E}\left(\max \left\\{X^{2}, Y^{2}\right\\}\right) \leq 1+\sqrt{1-\rho^{2}}\)

Three players, \(A, B\), and \(C\), take tums to roll a die; they do this in the order \(A B C A B C A . \ldots\) (a) Show that the probability that, of the three players, \(\mathrm{A}\) is the first to throw a \(6, \mathrm{~B}\) the second, and C the third, is \(216 / 1001\). (b) Show that the probability that the first 6 to appear is thrown by \(\mathrm{A}\), the second 6 to appear is thrown by \(\mathrm{B}\), and the third 6 to appear is thrown by \(\mathrm{C}\), is \(46656 / 753571\).

A coin is tossed repeatedly, heads turning up with probability \(p\) on each toss. Player \(\mathrm{A}\) wins the game if \(m\) heads appear before \(n\) tails have appeared, and player B wins otherwise. Let \(p_{m n}\) be the probability that \(A\) wins the game. Set up a difference equation for the \(p_{m n}\). What are the boundary conditions?

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