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Let \(p_{0}=P[X=0\\}\) and suppose that \(0

Short Answer

Expert verified
The short version of the answer is: (a) \(E[X \mid X \neq 0] = \frac{\mu}{1 - p_0}\) (b) \(\operatorname{Var}(X \mid X \neq 0) = \left(\frac{\sigma^2 + \mu^2}{1 - p_0}\right) - \left(\frac{\mu}{1 - p_0}\right)^2\)

Step by step solution

01

(a) Find the conditional expectation E[X | X ≠ 0]

To find the conditional expectation \(E[X \mid X \neq 0]\), we need to consider what the expected value of \(X\) would be if we ignored the cases where \(X = 0\). We can calculate this by using the definition of conditional expectation: \[E[X \mid X \neq 0] = \frac{E[X \cdot 1_{\{ X \neq 0\}}]}{P[X \neq 0]}\] where \(1_{\{ X \neq 0\}}\) is the indicator function that takes on the value 1 when \(X \neq 0\) and 0 otherwise. Notice that when \(X = 0\), the product \(X \cdot 1_{\{ X \neq 0\}}\) is also 0. Therefore, we can rewrite the conditional expectation as a ratio that only considers the non-zero values of \(X\): \[E[X \mid X \neq 0] = \frac{E[X]}{P[X \neq 0]}\] Since we know \(\mu = E[X]\) and \(p_0 = P[X = 0]\), we can easily find \(P[X \neq 0]\): \[P[X \neq 0] = 1 - P[X = 0] = 1 - p_0\] Finally, we can plug in our known values of \(\mu\) and \(1 - p_0\) to calculate the conditional expectation: \[E[X \mid X \neq 0] = \frac{\mu}{1 - p_0}\]
02

(b) Find the conditional variance Var(X | X ≠ 0)

Now let's find the conditional variance \(\operatorname{Var}(X \mid X \neq 0)\). We can make use of the property that relates conditional variance to conditional second moments: \[\operatorname{Var}(X \mid X \neq 0) = E[X^2 \mid X \neq 0] - (E[X \mid X \neq 0])^2\] We have already calculated \(E[X \mid X \neq 0]\) in part (a). What remains is to find \(E[X^2 \mid X \neq 0]\). Just like in the conditional expectation calculation, we can express the conditional second moment as a ratio involving the non-zero values of \(X\): \[E[X^2 \mid X \neq 0] = \frac{E[X^2 \cdot 1_{\{ X \neq 0\}}]}{P[X \neq 0]}\] We have already found \(P[X \neq 0]\) in part (a). Now we need to find the unconditional second moment \(E[X^2 \cdot 1_{\{ X \neq 0\}}]\). Notice that: \[E[X^2] = P[X=0]\cdot(0)^2 + E[X^2 \mid X \neq 0] \cdot P[X \neq 0]\] We have all the information we need to calculate \(E[X^2]\). Using the formula \(\sigma^2 = \operatorname{Var}(X) = E[X^2] - E[X]^2\), we can solve for \(E[X^2]\): \[E[X^2] = \sigma^2 + \mu^2\] Now we can plug in the values for \(E[X^2]\) and \(P[X \neq 0]\) to find \(E[X^2 \mid X \neq 0]\): \[E[X^2 \mid X \neq 0] = \frac{\sigma^2 + \mu^2}{1 - p_0}\] Finally, we can use our results for \(E[X \mid X \neq 0]\) and \(E[X^2 \mid X \neq 0]\) in the conditional variance formula: \[\operatorname{Var}(X \mid X \neq 0) = \left(\frac{\sigma^2 + \mu^2}{1 - p_0}\right) - \left(\frac{\mu}{1 - p_0}\right)^2\]

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Key Concepts

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

Variance
Variance is a measure of how much a set of numbers are spread out from their average value. In probability and statistics, it helps us understand the amount of variability in a random variable. The formula for variance is given by:
\[\operatorname{Var}(X) = E[(X - \mu)^2]\]
Where:
  • \(X\) is the random variable.
  • \(\mu\) is the mean or expected value of \(X\).
  • \(E\) denotes the expectation.
Variance is especially useful in financial mathematics and risk analysis, as it gives insight into the risk or uncertainty associated with different investments. A larger variance suggests higher risk, as values deviate more from the average, while a smaller variance indicates less risk.
Conditional Variance
Conditional variance is an extension of variance that applies when we consider a subset of possibilities, excluding certain conditions. It reflects how much the values of a random variable differ from the expected value, given some condition.
The formula for conditional variance is:
\[\operatorname{Var}(X \mid Y) = E[(X - E[X \mid Y])^2 \mid Y]\]
In our exercise, we calculate:
\[\operatorname{Var}(X \mid X eq 0)\]
This involves first finding the mean of \(X\) when \(X eq 0\), and then calculating how much the values deviate from this conditional mean. Understanding conditional variance helps in many fields like economics, where it's crucial to know the variance in income or asset prices excluding extreme scenarios.
Probability
Probability is a fundamental concept in statistics and mathematics, representing the likelihood that a particular event will occur. It's expressed as a number between 0 and 1, where 0 means the event will not happen, and 1 indicates certainty.
Key aspects of probability include:
  • The probability of an event \(A\), denoted \(P(A)\), represents how likely \(A\) is to occur.
  • If \(P(A) = 1\), the event is certain.
  • If \(P(A) = 0\), the event is impossible.
In this exercise, concepts of probability are used to determine conditional values, such as finding \(P[X eq 0]\), which is crucial for calculating expectations and variances under different conditions. Through probability, we quantify and manage uncertainty in our predictions.
Mathematical Expectation
Mathematical expectation, often called expected value, is a core idea in probability that represents the average outcome of a random variable if an experiment is repeated many times. It's calculated by:
\[E[X] = \sum_x x \, P(X = x)\]
Where:
  • \(x\) are the possible values of the random variable \(X\).
  • \(P(X = x)\) is the probability of \(X\) taking the value \(x\).
Expectation can be understood as what you "expect" on average in the long run. It plays an essential role in decision-making and risk assessment by averaging possible outcomes. In finance, for instance, it helps investors understand the average expected returns on an investment.

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Most popular questions from this chapter

Suppose that we want to predict the value of a random variable \(X\) by using one of the predictors \(Y_{1}, \ldots, Y_{n}\), each of which satisfies \(E\left[Y_{i} \mid X\right]=X .\) Show that the predictor \(Y_{i}\) that minimizes \(E\left[\left(Y_{i}-X\right)^{2}\right]\) is the one whose variance is smallest. Hint: Compute \(\operatorname{Var}\left(Y_{i}\right)\) by using the conditional variance formula.

If \(R_{i}\) denotes the random amount that is earned in period \(i\), then \(\sum_{i=1}^{\infty} \beta^{i-1} R_{i}\), where \(0<\beta<1\) is a specified constant, is called the total discounted reward with discount factor \(\beta .\) Let \(T\) be a geometric random variable with parameter \(1-\beta\) that is independent of the \(R_{i} .\) Show that the expected total discounted reward is equal to the expected total (undiscounted) reward earned by time \(T\). That is, show that $$ E\left[\sum_{i=1}^{\infty} \beta^{i-1} R_{i}\right]=E\left[\sum_{i=1}^{T} R_{i}\right] $$

A coin, having probability \(p\) of landing heads, is continually flipped until at least one head and one tail have been flipped. (a) Find the expected number of flips needed. (b) Find the expected number of flips that land on heads. (c) Find the expected number of flips that land on tails. (d) Repeat part (a) in the case where flipping is continued until a total of at least two heads and one tail have been flipped.

Show in the discrete case that if \(X\) and \(Y\) are independent, then $$ E[X \mid Y=y]=E[X] \text { for all } y $$

An individual whose level of exposure to a certain pathogen is \(x\) will contract the disease caused by this pathogen with probability \(P(x) .\) If the exposure level of a randomly chosen member of the population has probability density function \(f\), determine the conditional probability density of the exposure level of that member given that he or she (a) has the disease. (b) does not have the disease. (c) Show that when \(P(x)\) increases in \(x\), then the ratio of the density of part (a) to that of part (b) also increases in \(x\).

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