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Chapter 5: Problem 51

Let \(X\) and \(Y\), reaction times (sec) to two different stimuli, have a bivariate normal distribution with mean \(\mu_{1}=20\) and standard deviation \(\sigma_{1}=2\) for \(X\) and mean \(\mu_{2}=30\) and standard deviation \(\sigma_{2}=5\) for \(Y\). Assume \(\rho=.8\). Determine a. \(\mu_{Y \mid X=x}\) b. \(\sigma_{Y \mid X=x}^{2}\) c. \(\sigma_{Y \mid X=x}\) d. \(P(Y>46 \mid X=25)\)

Short Answer

Expert verified
a. 40 b. 9 c. 3 d. 0.0228

Step by step solution

01

Find Conditional Mean

The formula for the conditional mean \(\mu_{Y \mid X=x}\) of a bivariate normal distribution is:\[ \mu_{Y \mid X=x} = \mu_{2} + \rho \frac{\sigma_{2}}{\sigma_{1}}(x - \mu_{1}) \]Substituting the given values:\[ \mu_{Y \mid X=25} = 30 + 0.8 \times \frac{5}{2}(25 - 20) \]Calculate the expression:\[ = 30 + 0.8 \times 2.5 \times 5 = 30 + 10 = 40 \]Thus, \(\mu_{Y \mid X=25} = 40\).
02

Find Conditional Variance

The formula for the conditional variance \(\sigma_{Y \mid X=x}^{2}\) is:\[ \sigma_{Y \mid X=x}^{2} = \sigma_{2}^{2}(1 - \rho^{2}) \]Substituting the given values:\[ \sigma_{Y \mid X=25}^{2} = 5^2(1 - 0.8^2) \]Calculate the expression:\[ = 25 \times (1 - 0.64) = 25 \times 0.36 = 9 \]Thus, \(\sigma_{Y \mid X=25}^{2} = 9\).
03

Find Conditional Standard Deviation

To find \(\sigma_{Y \mid X=x}\), take the square root of the conditional variance found in the previous step:\[ \sigma_{Y \mid X=25} = \sqrt{9} = 3 \]Thus, \(\sigma_{Y \mid X=25} = 3\).
04

Calculate Conditional Probability

We need to find \(P(Y>46 \mid X=25)\). First, we standardize using the conditional mean and standard deviation:\[ Z = \frac{Y - \mu_{Y \mid X=x}}{\sigma_{Y \mid X=x}} = \frac{46 - 40}{3} = 2 \]The probability we want is \(P(Z > 2)\), which can be found using the standard normal table or a calculator:\[ P(Z > 2) \approx 0.0228 \]Thus, \(P(Y>46 \mid X=25) \approx 0.0228\).

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Key Concepts

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

Conditional Mean
The conditional mean of a bivariate normal distribution tells us the expected value of one variable given the presence of another. It's like saying, "Knowing that one variable has happened, what should we expect from the other?" In math terms, if we know the value of \( X \), the conditional mean \( \mu_{Y \mid X=x} \) helps us predict \( Y \).

The formula is:
  • \( \mu_{Y \mid X=x} = \mu_{2} + \rho \frac{\sigma_{2}}{\sigma_{1}}(x - \mu_{1}) \)
To break it down simply:
  • \( \mu_{2} \) is the mean of \( Y \) when \( X \) doesn't matter.
  • \( \rho \) is the correlation; how much \( X \) and \( Y \) are related.
  • \( \frac{\sigma_{2}}{\sigma_{1}} \) adjusts for the different standard deviations of \( Y \) and \( X \).
  • \( x - \mu_{1} \) tells us how different \( X \) is from its average, adjusting our guess for \( Y \).
In our exercise, if \( X = 25 \), the conditional mean for \( Y \) becomes 40, showing how the expected reaction time for \( Y \) changes based on knowing \( X \).
Conditional Variance
Conditional variance measures how much we expect the value of \( Y \) to vary when we're already given a specific \( X \). It's like saying, "Given this starting point, how uncertain are we about where \( Y \) might land?"

The formula to calculate conditional variance is:
  • \( \sigma_{Y \mid X=x}^{2} = \sigma_{2}^{2}(1 - \rho^{2}) \)
Here's what it signifies:
  • \( \sigma_{2}^{2} \) is the original variance of \( Y \), showing how much \( Y \) naturally varies.
  • \( 1 - \rho^{2} \) decreases this variance based on how \( X \) and \( Y \) are connected. The stronger the relationship (higher \( \rho \)), the less uncertainty there is when we know \( X \).
In our problem, since \( \rho = 0.8 \), our conditional variance for \( Y \) ends up as 9, indicating the new level of variation expected with \( X \) known.
Conditional Probability
Conditional probability tells us the chances of an event happening given another event has already occurred. It's like asking, "What's the probability of \( Y \) doing something, given \( X \) already happened?"

The formal expression involves adopting a normal distribution approach by using the known mean and standard deviation. We standardize it with:
  • \( Z = \frac{Y - \mu_{Y \mid X=x}}{\sigma_{Y \mid X=x}} \)
The idea is to convert \( Y \) into a standard normal variable \( Z \). This compares our situation against the familiar normal distribution curve:
  • In our case, if \( Y > 46 \), it turns into finding \( P(Z > 2) \).
Using normal distribution tables or calculators tells us this probability is around 0.0228. That means there鈥檚 about a 2.28% chance for this less-likely event, given our condition.
Standard Normal Distribution
The standard normal distribution is a regular bell-shaped curve that's been standardized for easier calculation. This standard form has a mean of 0 and a standard deviation of 1, making calculations of probabilities straightforward.

A standard normal distribution comes into play when we transform any other normal distribution into this format by using the Z score calculation:
  • \( Z = \frac{X - \mu}{\sigma} \)
This step is essential in understanding probabilities or unusual events in data:
  • If you have a Z score, it tells you how far away a value is from the mean, measured in standard deviations.
  • This process helped us find the probability that \( Y \) was greater than 46 in our prior example, using the value 2 from the Z calculation.
A Z score like 2 indicates the event is two standard deviations above the mean, highlighting a less common occurrence.

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Most popular questions from this chapter

A circular sampling region with radius \(X\) is chosen by a biologist, where \(X\) has an exponential distribution with mean value \(10 \mathrm{ft}\). Plants of a certain type occur in this region according to a (spatial) Poisson process with "rate" \(.5\) plant per square foot. Let \(Y\) denote the number of plants in the region. a. Find \(E(Y \mid X=x)\) and \(V(Y \mid X=x)\) b. Use part (a) to find \(E(Y)\). c. Use part (a) to find \(V(Y)\).

Consider a sealed-bid auction in which each of the \(n\) bidders has his/her valuation (assessment of inherent worth) of the item being auctioned. The valuation of any particular bidder is not known to the other bidders. Suppose these valuations constitute a random sample \(X_{1}, \ldots, X_{n}\) from a distribution with cdf \(F(x)\), with corresponding order statistics \(Y_{1} \leq Y_{2} \leq \cdots \leq Y_{n}\). The rent of the winning bidder is the difference between the winner's valuation and the price. The article "Mean Sample Spacings, Sample Size and Variability in an Auction-Theoretic Framework" (Oper. Res. Lett., 2004: 103-108) argues that the rent is just \(Y_{n}-Y_{n-1}\) (why?) a. Suppose that the valuation distribution is uniform on \([0,100]\). What is the expected rent when there are \(n=10\) bidders? b. Referring back to (a), what happens when there are 11 bidders? More generally, what is the relationship between the expected rent for \(n\) bidders and for \(n+1\) bidders? Is this intuitive?

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A pizza place has two phones. On each phone the waiting time until the first call is exponentially distributed with mean one minute. Each phone is not influenced by the other. Let \(X\) be the shorter of the two waiting times and let \(Y\) be the longer. It can be shown that the joint pdf of \(X\) and \(Y\) is \(f(x, y)=2 e^{-(x+y)}, 0

A market has both an express checkout line and a superexpress checkout line. Let \(X_{1}\) denote the number of customers in line at the express checkout at a particular time of day, and let \(X_{2}\) denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of \(X_{1}\) and \(X_{2}\) is as given in the accompanying table. $$ \begin{array}{cc|cccc} & & & &{x_{2}} & \\ & & 0 & 1 & 2 & 3 \\ \hline & 0 & .08 & .07 & .04 & .00 \\ & 1 & .06 & .15 & .05 & .04 \\ x_{1} & 2 & .05 & .04 & .10 & .06 \\ & 3 & .00 & .03 & .04 & .07 \\ & 4 & .00 & .01 & .05 & .06 \end{array} $$ a. What is \(P\left(X_{1}=1, X_{2}=1\right)\), that is, the probability that there is exactly one customer in each line? b. What is \(P\left(X_{1}=X_{2}\right)\), that is, the probability that the numbers of customers in the two lines are identical? c. Let \(A\) denote the event that there are at least two more customers in one line than in the other line. Express \(A\) in terms of \(X_{1}\) and \(X_{2}\), and calculate the probability of this event. d. What is the probability that the total number of customers in the two lines is exactly four? At least four? e. Determine the marginal pmf of \(X_{1}\), and then calculate the expected number of customers in line at the express checkout. f. Determine the marginal pmf of \(X_{2}\). g. By inspection of the probabilities \(P\left(X_{1}=4\right)\), \(P\left(X_{2}=0\right)\), and \(P\left(X_{1}=4, X_{2}=0\right)\), are \(X_{1}\) and \(X_{2}\) independent random variables? Explain.
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