Problem 66
Question
The logistic curve \(F(y)=\frac{1}{1+e^{-y}},-\infty
Step-by-Step Solution
Verified Answer
The three conditions for the logistic curve to represent a cdf were successfully verified: 1) \(\lim _{y \rightarrow-\infty} \frac{1}{1+e^{-y}}=0\), 2) \(\lim _{y \rightarrow+\infty} \frac{1}{1+e^{-y}}=1\), and 3) the function is strictly increasing. The associated pdf is given by \(f(y) = \frac{e^{-y}}{(1+e^{-y})^2}\)
1Step 1: Verification of Condition 1
To verify the first condition let's find the limit of the given function as \(y\) approaches \(-\infty\). According to limit properties, when \(e^{-y}\) approaches \(-\infty\), \(e^{-y}\) approaches infinity (\(e^{-y}=e^{+\infty}=+\infty\)). So the expression becomes \(F(y)=\frac{1}{1+e^{-y}} = \frac{1}{1+\infty} = 0\). Thus, the first condition is verified.
2Step 2: Verification of Condition 2
To verify the second condition let's find the limit of the function as \(y\) approaches \(+\infty\). According to limit properties, when \(e^{-y}\) approaches \(+\infty\), \(e^{-y}\) approaches 0 (\(e^{-y}=e^{-\infty}= 0\)). So the expression becomes \(F(y)=\frac{1}{1+e^{-y}} = \frac{1}{1+0} = 1\). Thus, the second condition is verified.
3Step 3: Verification of Condition 3
For this condition, inspection of the logistic curve is a simple method to verify it is strictly increasing. If the derivative of \(F(y)\) is positive, this confirms that indeed the function is increasing. Let's find the derivative to verify it: \(F'(y)=e^{-y}(1+e^{-y})^{-2}\). On elaboration, \(F'(y)= \frac{e^{-y}}{(1+e^{-y})^2}\), which is always greater than 0, for all values of \(y\), thus the function is strictly increasing.
4Step 4: Finding the Associated PDF
By definition, the pdf is the derivative of the cdf. Therefore to find the associated pdf, we find the derivative of the given function \(F(y)=\frac{1}{1+e^{-y}}\). Let's write \(F(y)\) in a more convenient way: \(F(y) = (1 + e^{-y})^{-1}\). Differentiating, we get that the pdf \(f(y) = F'(y)= \frac{e^{-y}}{(1+e^{-y})^2}\)
Key Concepts
Cumulative Distribution Function (CDF)Probability Density Function (PDF)Derivatives
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function, commonly abbreviated as CDF, is crucial in probability and statistics. It describes the probability that a continuous random variable will have a value less than or equal to a specific point. For any random variable with a logistic distribution, such as the logistic curve provided: \(F(y)=\frac{1}{1+e^{-y}}\), the CDF gives us critical insights into the distribution's behavior.
When examining the logistic curve, we know it is `increasing`. This means as \(y\) increases, the probability \(F(y)\) also increases. In the context of our logistic curve formula, ensure it fits three specific criteria:
When examining the logistic curve, we know it is `increasing`. This means as \(y\) increases, the probability \(F(y)\) also increases. In the context of our logistic curve formula, ensure it fits three specific criteria:
- As \(y\) approaches \(-\infty\), \(F(y)\) should approach 0.
- As \(y\) approaches \(+\infty\), \(F(y)\) should approach 1.
- The curve itself should always rise, never fall, indicating an ascending CDF.
Probability Density Function (PDF)
The Probability Density Function, or PDF, is a fundamental concept that describes how the probability of a random variable is distributed. The PDF is the derivative of the CDF, providing the likelihood of the random variable taking on an exact value. For continuous variables, this notion can be linked to the slope of the distribution. In our logistic example, the PDF is derived by taking the derivative of the CDF:\[ f(y) = F'(y) = \frac{e^{-y}}{(1 + e^{-y})^2} \]This function represents the PDF of the logistic distribution. The PDF is essential because it reveals where values are more dense or sparse in the distribution. When the PDF is high, we are more likely to observe values near that part of the distribution.
By understanding the PDF, you'd ascertain the aspects of the distribution such as the mean, variance, and skewness, essential for statistical analysis and applications in various fields such as finance and biology.
By understanding the PDF, you'd ascertain the aspects of the distribution such as the mean, variance, and skewness, essential for statistical analysis and applications in various fields such as finance and biology.
Derivatives
Derivatives play a key role in understanding and analyzing functions in mathematics and statistics, elucidating how a function changes at any point. In the context of probability distributions, derivatives help transition from a cumulative perspective (CDF) to a density perspective (PDF).
For instance, given the logistic distribution's CDF: \(F(y)=\frac{1}{1+e^{-y}}\), finding its derivative allows us to outline the PDF. The process involves applying basic derivative rules and, sometimes, the chain rule to achieve an appropriate form. For the logistic curve:
For instance, given the logistic distribution's CDF: \(F(y)=\frac{1}{1+e^{-y}}\), finding its derivative allows us to outline the PDF. The process involves applying basic derivative rules and, sometimes, the chain rule to achieve an appropriate form. For the logistic curve:
- Differentiating \((1 + e^{-y})^{-1}\) using the chain rule results in \( F'(y) = \frac{e^{-y}}{(1 + e^{-y})^2} \).
- The positive derivative implies the CDF is strictly increasing, proving the function's validity as a CDF.
Other exercises in this chapter
Problem 64
Suppose \(F_{Y}(y)=\frac{1}{12}\left(y^{2}+y^{3}\right), 0 \leq y \leq 2\). Find \(f_{Y}(y)\).
View solution Problem 65
In a certain country, the distribution of a family's disposable income, \(Y\), is described by the pdf \(f_{Y}(y)=\) \(y e^{-y}, y \geq 0\). Find \(F_{Y}(y)\).
View solution Problem 68
Suppose that \(f_{Y}(y)\) is a continuous and sym. pdf, where symmetry is the property that \(f_{Y}(y)=f\) for all \(y\). Show that \(P(-a \leq Y \leq a)=2 F_{Y
View solution Problem 70
Recall the game of Keno described in Question 3.2.26. The following are all the payoffs on a \(\$ 1\) wager where the player has bet on ten numbers. Calculate \
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