Problem 152
Question
An article in Mathematical Biosciences ["Influence of Delayed Viral Production on Viral Dynamics in HIV-1 Infected Patients" \((1998,\) Vol. \(152(2),\) pp. \(143-163)]\) considered the time delay between the initial infection by immunodeficiency virus type 1 (HIV-1) and the formation of productively infected cells. In the simulation model, the time delay is approximated by a gamma distribution with parameters \(r=4\) and \(1 / \lambda=0.25\) days. Determine the following: (a) Mean and variance of time delay (b) Probability that a time delay is more than half a day (c) Probability that a time delay is between one-half and one day
Step-by-Step Solution
Verified Answer
(a) Mean: 1 day, Variance: 0.25 days²; (b) Calculate using gamma CDF; (c) Calculate using gamma CDF.
1Step 1: Understand the Gamma Distribution
The gamma distribution is defined by two parameters: the shape parameter \(r\) and the rate parameter \(\lambda\). The article provides these as \(r=4\) and \(\frac{1}{\lambda} = 0.25\) days. Thus, the rate \(\lambda\) is \(\frac{1}{0.25} = 4\).
2Step 2: Calculate Mean and Variance
For the gamma distribution, the mean \(\mu\) is given by \(\frac{r}{\lambda}\) and the variance \(\sigma^2\) is given by \(\frac{r}{\lambda^2}\). Using the given parameters: - Mean \(\mu = \frac{4}{4} = 1\) day - Variance \(\sigma^2 = \frac{4}{4^2} = \frac{4}{16} = 0.25\) days².
3Step 3: Calculate Probability for More than Half a Day
For the gamma distribution, the cumulative distribution function (CDF) can be used. We need \(P(X > 0.5) = 1 - P(X \leq 0.5)\). Using a standard gamma distribution table or calculator for \(P(X \leq 0.5)\), compute this probability, given a gamma distribution with \(r=4\) and \(\lambda=4\).
4Step 4: Calculate Probability Between Half and One Day
We want \(P(0.5 < X < 1) = P(X < 1) - P(X \leq 0.5)\). Using a gamma distribution table or software, find \(P(X < 1)\) and use the previously computed \(P(X \leq 0.5)\). Subtract to obtain the desired probability.
Key Concepts
Probability CalculationMean and VarianceCumulative Distribution Function
Probability Calculation
When working with the Gamma Distribution, calculating probabilities involves understanding its cumulative nature. Specifically, we utilize the cumulative distribution function (CDF), which helps us determine the probability that a random variable is less than or equal to a certain value. In the given problem, we are interested in the probabilities concerning specific time delays modeled by the Gamma Distribution.
### Calculating Specific ProbabilitiesTo find the probability of a time delay being more than half a day, we use:
Understanding the CDF in the Gamma Distribution context allows us to evaluate these probabilities effectively, ensuring we can describe the likelihood of specific events in a real-world scenario.
### Calculating Specific ProbabilitiesTo find the probability of a time delay being more than half a day, we use:
- 1 minus the CDF for 0.5 days. This probability is expressed as \(P(X > 0.5) = 1 - P(X \leq 0.5)\).
- The CDF for 1 day minus the CDF for 0.5 days: \(P(0.5 < X < 1) = P(X < 1) - P(X \leq 0.5)\).
Understanding the CDF in the Gamma Distribution context allows us to evaluate these probabilities effectively, ensuring we can describe the likelihood of specific events in a real-world scenario.
Mean and Variance
Mean and variance are core concepts when working with any probability distribution, including the gamma distribution. They provide crucial insights into the distribution's behavior, such as its central tendency and the spread of possible values. For our exercise, parameters play an essential role in determining these characteristics.
### Calculating the MeanFor the gamma distribution, the mean \( \mu \) is derived from its parameters by the formula:
### Calculating the VarianceVariance \( \sigma^2 \) is another metric providing insights into the spread of the distribution. It's calculated by:
### Calculating the MeanFor the gamma distribution, the mean \( \mu \) is derived from its parameters by the formula:
- \( \mu = \frac{r}{\lambda} \)
- \( \mu = \frac{4}{4} = 1 \) day.
### Calculating the VarianceVariance \( \sigma^2 \) is another metric providing insights into the spread of the distribution. It's calculated by:
- \( \sigma^2 = \frac{r}{\lambda^2} \)
- \( \sigma^2 = \frac{4}{4^2} = \frac{4}{16} = 0.25 \) days².
Cumulative Distribution Function
The cumulative distribution function (CDF) is an essential tool when working with the gamma distribution to understand the cumulative probability up to a certain threshold. It helps us compute the probability of a variable being less than or equal to a specific value.
### What is the CDF?The CDF for a random variable \(X\) gives us the probability that \(X\) will take a value less than or equal to \(x\). Mathematically, it is defined as:
### Utilizing the CDF in the Gamma DistributionFor this particular type of distribution, calculating probabilities involves using the CDF to find the difference between two probabilities:\ \( P(a < X < b) = F(b) - F(a) \). For example, in our exercise, to find the probability between 0.5 and 1 day, we compute:\ \( P(0.5 < X < 1) = F(1) - F(0.5) \).
This process often requires statistical tables or computational methods, as exact mathematical expressions for the CDF in case of the gamma distribution for non-standard values need complex integration.
### What is the CDF?The CDF for a random variable \(X\) gives us the probability that \(X\) will take a value less than or equal to \(x\). Mathematically, it is defined as:
- \( F(x) = P(X \leq x) \)
### Utilizing the CDF in the Gamma DistributionFor this particular type of distribution, calculating probabilities involves using the CDF to find the difference between two probabilities:\ \( P(a < X < b) = F(b) - F(a) \). For example, in our exercise, to find the probability between 0.5 and 1 day, we compute:\ \( P(0.5 < X < 1) = F(1) - F(0.5) \).
This process often requires statistical tables or computational methods, as exact mathematical expressions for the CDF in case of the gamma distribution for non-standard values need complex integration.
Other exercises in this chapter
Problem 149
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View solution Problem 150
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View solution Problem 153
Suppose that \(X\) has a Weibull distribution with \(\beta=0.2\) and \(\delta=100\) hours. Determine the mean and variance of \(X\).
View solution Problem 154
Suppose that \(X\) has a Weibull distribution with \(\beta=0.2\) and \(\delta=100\) hours. Determine the following: (a) \(P(X5000)\)
View solution