Problem 4 A certain community is composed ... [FREE SOLUTION]

Chapter 4: Problem 4

A certain community is composed of $m$ families, $n_{i}$ of which have $i$ children, $\sum_{i=1}^{r} n_{i}=m .$ If one of the families is randomly chosen, let $X$ denote the number of children in that family. If one of the $\sum_{i=1}^{r}$ in $_{l}$ children is randomly chosen, let $Y$ denote the total number of children in the family of that child. Show that $E[Y] \geq E[X]$.

Short Answer

Expert verified

The expected number of children in a family when choosing a child randomly ($E[Y]$) is greater than or equal to the expected number of children in a family when choosing a family randomly ($E[X]$), as shown below: $$ E[Y] = \sum_{i=1}^r i * \frac{i*n_i}{\sum_{i=1}^r i*n_i} \geq \sum_{i=1}^r i * \frac{n_i}{m} = E[X] $$

Step by step solution

Probability Mass Function of X

To find the PMF of $X$, we need to compute the probability that a randomly chosen family has $i$ children, for $i=1,2,...,r$. $$P(X=i) = \frac{n_i}{m}$$

Expectation of X

Now we can find the expected value of $X$, using the definition of expectation. The expectation of the number of children in a family when a family is chosen randomly is: $$E[X] = \sum_{i=1}^r i * P(X=i) = \sum_{i=1}^r i * \frac{n_i}{m}$$

Probability Mass Function of Y

To find the PMF of $Y$, we need to compute the probability that a randomly chosen child comes from a family with $i$ children, for $i=1,2,...,r$. $$P(Y=i) = \frac{i*n_i}{\sum_{i=1}^r i*n_i}$$

Expectation of Y

Now we can find the expected value of $Y$, using the definition of expectation. The expectation of the number of children in a family when a child is chosen randomly is: $$E[Y] = \sum_{i=1}^r i * P(Y=i) = \sum_{i=1}^r i * \frac{i*n_i}{\sum_{i=1}^r i*n_i}$$

Comparing E[Y] and E[X]

Let's compare the expressions we found for $E[Y]$ and $E[X]$: $$ \begin{aligned} E[Y] &= \sum_{i=1}^r i * \frac{i*n_i}{\sum_{i=1}^r i*n_i} \\ &E[X] = \sum_{i=1}^r i * \frac{n_i}{m} \end{aligned} $$ Now, let's multiply both sides by $m * \sum_{i=1}^r i*n_i$ to cancel out the denominators: $$ \begin{aligned} m * \sum_{i=1}^r i * \frac{i*n_i}{\sum_{i=1}^r i*n_i} &\geq m*\sum_{i=1}^r i * \frac{n_i}{m} \\ m * \sum_{i=1}^r i^2 * n_i &\geq \sum_{i=1}^r i*n_i^2 \end{aligned} $$ Observe that $m * \sum_{i=1}^r i^2 * n_i$ is the sum of squares of all the children in the families weighted by the total number of families ($m$), while $\sum_{i=1}^r i*n_i^2$ is the sum of squares of all the children in the families weighted by the number of families with that particular number of children in each respective family. Since for all values of $i \geq 1$, $i^2 \geq i$ and hence also $m * i^2 \geq i * n_i$ as $m * i^2$ takes the sum over all families, we can conclude that: $$ m * \sum_{i=1}^r i^2 * n_i \geq \sum_{i=1}^r i*n_i^2 $$ Therefore, we have shown that $E[Y] \geq E[X]$. This means that the expected number of children in a family when choosing a child randomly is greater than or equal to the expected number of children in a family when choosing a family randomly.

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Key Concepts

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

Probability Mass Function

Understanding the probability mass function (PMF) is crucial when dealing with discrete random variables. In simple terms, a PMF assigns a probability to each possible outcome of a random variable. For instance, if we're tossing a fair coin, the PMF would assign a probability of 0.5 to both getting a 'heads' and getting a 'tails'.

In the family size distribution context, the PMF helps us calculate the likelihood that a randomly chosen family has a certain number of children. To compute this, we divide the number of families with a particular number of children by the total number of families. This gives us a probability distribution over the family sizes, which is essential for later determining the expected number of children in a randomly selected family or from a child-centric perspective.

Random Variable Expectation

The expectation of a random variable, often symbolized as E[X], is the long-term average value of repetitions of the experiment it represents. It can be thought of as the probability-weighted average of all possible outcomes. For a random variable representing the number of children in a randomly selected family, the expectation is calculated by summing the products of each family size (number of children) and the probability of a family having that size. It's like calculating the average family size, but instead of each family contributing equally to the average, families with more members (children) have more 'weight' in this calculation.

This concept of weighting is the key to understanding why the expected number of children in a family when picking a child (E[Y]) may differ from simply picking a family (E[X]), as larger families have a greater chance of being selected when starting from the child's perspective.

Combinatorics in Probability

Combinatorics, the mathematics of counting, plays a fundamental role in probability, especially when determining the number of ways certain events can occur. In the probability realm, it enables us to figure out the number of possible outcomes that satisfy specific conditions. For instance, if we're looking at the number of ways to form a family with a given number of children from a larger pool, combinatorial methods would be used.

In our family size example, combinatorics isn't used explicitly, but it underpins the reasoning why larger families are more likely to be chosen when selecting a child at random. This is because there are more 'combinations' of children from larger families than from smaller ones, influencing the probability mass function of Y.

Family Size Distribution

The distribution of family sizes within a population can be represented by a frequency distribution, showing how families are distributed across different sizes (number of children). In our context, we explore two types of averages within this distribution: one where we pick families at random, and one where we pick children at random.

The key takeaway is the 'larger family bias' that occurs when picking a child at random鈥攐ne is more likely to pick a child from a large family than a small family because there are more children from large families to choose from. This result, E[Y] 鈮� E[X], is a direct consequence of how the random variable expectation changes depending on whether we approach the problem by looking at families as units or individual children within those families as our starting point.

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Short Answer

Step by step solution

Probability Mass Function of X

Expectation of X

Probability Mass Function of Y

Expectation of Y

Comparing E[Y] and E[X]

Key Concepts

Probability Mass Function

Random Variable Expectation

Combinatorics in Probability

Family Size Distribution

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