Problem 267
Question
The form of the moment-generating function for a normal random variable is \(M_{Y}(t)=e^{a t t+b^{2} t^{2} / 2}\) (recall Example 3.12.4). Differentiate \(M_{Y}(t)\) to verify that \(a=E(Y)\) and \(b^{2}=\operatorname{Var}(Y)\)
Step-by-Step Solution
Verified Answer
By differentiating the Moment-Generating Function of a normally distributed random variable, we can establish that \( a = E(Y) \) and \( b^2 = Var(Y) \).
1Step 1: Differentiate the Moment-Generating Function
We need to calculate the first and second derivatives of the Moment-Generating Function to derive the mean (\(a\)) and variance (\(b^2\)). For a normal distribution \(Y\), the Moment-Generating Function is \(M_{Y}(t)=e^{a t t+b^{2} t^{2} / 2}\). The first derivative of \(M_{Y}(t)\) with respect to \(t\) is \(M'_{Y}(t)= e^{at + 0.5b^{2}t^{2}}(at+b^{2}t)\). Substituting \(t=0\) in \(M'_{Y}(t)\), we will get \(a=E(Y)\) since by the property of MGF, the first derivative evaluated at 0 gives the expected value.
2Step 2: Calculate the second derivative
The second derivative of \(M_{Y}(t)\) with respect to \(t\) is \(M''_{Y}(t)= e^{at + 0.5b^{2}t^{2}}(at+b^{2}t)^{2} + e^{at + 0.5b^{2}t^{2}}(a+b^{2}t)\). Substituting \(t=0\) in \(M''_{Y}(t)\), we get \(a^{2}+b^{2}=Var(Y)\) since by the property of MGF, the second derivative evaluated at 0 gives the variance.
3Step 3: Derive variance from second derivative
We know that variance is equal to \(E(Y^{2}) - [E(Y)]^{2}\). Since \(a=E(Y)\), substituting \(a\) in \(a^{2}+b^{2}=Var(Y)\), we get \(b^{2}=Var(Y)\). Therefore, we have successfully verified that \(b^{2}=\operatorname{Var}(Y)\).
Key Concepts
Normal DistributionExpected ValueVariance
Normal Distribution
A normal distribution is a continuous probability distribution characterized by its symmetric, bell-shaped curve. It is a fundamental concept in statistics and is known for having specific mathematical properties.
In a normal distribution:
Its usefulness extends across various scientific fields due to its natural appearance in random processes, popularly known as the "bell curve."
In a normal distribution:
- The mean, median, and mode are all identical and lie at the center of the distribution.
- The distribution is symmetric about its mean.
- The spread of the curve is determined by the standard deviation, where about 68% of the data falls within one standard deviation from the mean, about 95% within two, and about 99.7% within three.
Its usefulness extends across various scientific fields due to its natural appearance in random processes, popularly known as the "bell curve."
Expected Value
The expected value is a fundamental concept in probability and statistics and refers to the mean or average of a random variable over a large number of experiments or trials. For a random variable \(Y\) with itself being normal, the expected value \(E(Y)\) represents the center of the distribution.
Using the moment-generating function (MGF), we can calculate the expected value:
Using the moment-generating function (MGF), we can calculate the expected value:
- The MGF of a random variable is given by the equation \(M_Y(t) = E(e^{tY})\).
- For a normal distribution, the MGF is \(M_Y(t) = e^{\mu t + \sigma^2 t^2 / 2}\).
- By differentiating this function with respect to \(t\), and evaluating it at \(t = 0\), we find that \(E(Y) = \mu\).
Variance
Variance measures how far a set of numbers is spread out from their average value. In the context of a normal distribution, variance quantifies the spread, or dispersion, of the data points around the mean.
For a random variable \(Y\) with a normal distribution, the variance \(Var(Y)\) is directly linked to the standard deviation \(\sigma\) by the equation: \[ Var(Y) = \sigma^2 \]The significance of variance includes the following:
For a random variable \(Y\) with a normal distribution, the variance \(Var(Y)\) is directly linked to the standard deviation \(\sigma\) by the equation: \[ Var(Y) = \sigma^2 \]The significance of variance includes the following:
- High variance indicates data points are more spread out from the mean.
- Low variance suggests data points are clustered closely around the mean.
- The second derivative gives \(a^2 + b^2 = Var(Y)\).
- Since \(a = E(Y)\), rearranging terms will verify \(b^2 = Var(Y)\).
Other exercises in this chapter
Problem 265
Calculate \(E\left(Y^{3}\right)\) for a random variable whose moment-generating function is \(M_{Y}(t)=e^{t^{2} / 2}\).
View solution Problem 266
Find \(E\left(Y^{4}\right)\) if \(Y\) is an exponential random variable with \(f_{Y}(y)=\lambda e^{-\lambda y}, y>0\).
View solution Problem 268
What is \(E\left(Y^{4}\right)\) if the random variable \(Y\) has moment-generating function \(M_{Y}(t)=(1-\alpha t)^{-k}\) ?
View solution Problem 270
Find an expression for \(E\left(Y^{k}\right)\) if \(M_{Y}(t)=(1-\) \(t / \lambda)^{-r}\), where \(\lambda\) is any positive real number and \(r\) is a positive
View solution