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If \(M_{X}(t)=1 /\left(1-t^{2}\right)\), find \(E(X)\) and \(V(X)\) by differentiating \(M_{X}(t)\).

Short Answer

Expert verified
The mean is 0 and the variance is 2.

Step by step solution

01

Understanding Moment Generating Function

The given moment generating function (MGF) is \(M_X(t) = \frac{1}{1-t^2}\). The moment generating function helps us find the moments of the random variable. The first and second derivatives of the MGF evaluated at \(t = 0\) give us the mean \(E(X)\) and the second moment \(E(X^2)\), which are used to determine the variance \(V(X)\).
02

Calculate First Derivative

Differentiate the MGF \(M_X(t) = \frac{1}{1-t^2}\) with respect to \(t\). The first derivative is found using the chain rule: \(M'_X(t) = \frac{2t}{(1-t^2)^2}\).
03

Evaluate First Derivative at t = 0

To find the mean \(E(X)\), substitute \(t = 0\) into the first derivative: \(M'_X(0) = \frac{2(0)}{(1-0^2)^2} = 0\). Thus, \(E(X) = 0\).
04

Calculate Second Derivative

Differentiate the first derivative \(M'_X(t) = \frac{2t}{(1-t^2)^2}\) again with respect to \(t\) to find the second derivative: \(M''_X(t) = \frac{2(1+t^2)}{(1-t^2)^3}\).
05

Evaluate Second Derivative at t = 0

To find \(E(X^2)\), substitute \(t = 0\) into the second derivative: \(M''_X(0) = \frac{2(1+0^2)}{(1-0^2)^3} = 2\). Thus, \(E(X^2) = 2\).
06

Calculate Variance

The variance \(V(X)\) is calculated using the formula \(V(X) = E(X^2) - [E(X)]^2\). Substitute the values: \(V(X) = 2 - (0)^2 = 2\).

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Key Concepts

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

Mean of a Random Variable
The mean of a random variable is essentially its expected value. It tells us the 'average' outcome we should expect from a random process. Think of it as the center of the probability distribution.
This average is calculated by summing up all possible values the random variable can take, each weighted by their probability of occurring.
For a continuous random variable, the mean is calculated by integrating the product of the variable and its probability density function over its entire range.
  • If you think about rolling a fair die, the mean would be the average number obtained from infinite rolls.
  • If the die has numbers 1 through 6, the mean is \[E(X) = \frac{1+2+3+4+5+6}{6} = 3.5\]
Quantifying this, a moment generating function (MGF) makes finding the mean straightforward. Evaluate the first derivative of the MGF at zero to find the mean: \[E(X) = M'_X(0)\]In our exercise, this results in finding that the mean, \(E(X)\), is 0.
Variance of a Random Variable
Variance gives us a measure of how much the values of the random variable spread out from the mean. Higher variance means data points are more spread out.
This concept is crucial for understanding the risk or uncertainty associated with the random variable's outcomes.
To calculate the variance, we use the formula:\[V(X) = E(X^2) - [E(X)]^2\]
  • \(E(X^2)\) is the expected value of the square of the random variable and tells us about the average of its squared deviations.
  • \([E(X)]^2\) is simply the square of the mean.
  • The difference between these two captures the spread of the distribution.
In this context, the MGF helps us by giving insights into \(E(X^2)\) from the second derivative. Like for our problem: the second derivative at zero shows \(E(X^2) = 2\).
Thus, the variance for our random variable is computed as \(V(X) = 2 - (0)^2 = 2\), indicating some level of spread around the mean.
Derivatives in Statistics
Derivatives in statistics are powerful tools for analyzing how a function changes. They are particularly useful in the context of moment generating functions to find expectations and variances.
The process involves using calculus, specifically differentiation, to understand the behavior of functions that represent statistical measures.
  • The first derivative of the MGF, taken with respect to \(t\), relates to the mean or expected value of a random variable.
  • The second derivative provides direct information about the spread or variance.
  • Each higher-order derivative correlates with higher-order moments which explain deeper properties of the distribution.
Applying this knowledge to our exercise, the first derivative gives us the zero mean value, and the second derivative helps us confirm the variance.
This not only solidifies our understanding of the particular random variable but also builds a conceptual framework to tackle a wide range of statistical problems efficiently.

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