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If you toss a fair die with outcome \(X, p(x)=\frac{1}{6}\) for \(x=1,2,3,4,5,6\). Find \(M_{x}(t)\).

Short Answer

Expert verified
The MGF is \(M_X(t) = \frac{1}{6}(e^t + e^{2t} + e^{3t} + e^{4t} + e^{5t} + e^{6t}).\)

Step by step solution

01

Define the Moment Generating Function

The moment-generating function (MGF) of a random variable X, denoted as \(M_X(t)\), is defined by the expected value \(E[e^{tX}]\). For a discrete random variable, the MGF is the sum of \(p(x) e^{tx}\) over all possible outcomes x.
02

Substitute the Probability Distribution Function

For a fair die, the probability of each outcome (1 through 6) is \(\frac{1}{6}\). Thus, we can write the MGF as \[ M_X(t) = E[e^{tX}] = \sum_{x=1}^{6} \frac{1}{6} e^{tx}. \]
03

Calculate the MGF for Each Outcome

Calculate for each possible outcome: \(e^{t \cdot 1}, e^{t \cdot 2}, \, ... \, e^{t \cdot 6}\). Then substitute these into the MGF formula: \[ M_X(t) = \frac{1}{6}(e^{t \cdot 1} + e^{t \cdot 2} + e^{t \cdot 3} + e^{t \cdot 4} + e^{t \cdot 5} + e^{t \cdot 6}). \]
04

Simplify the Expression

Factor out the common \(\frac{1}{6}\) and write the sum as: \[ M_X(t) = \frac{1}{6}(e^t + e^{2t} + e^{3t} + e^{4t} + e^{5t} + e^{6t}). \] This simplified expression represents the MGF of the random variable X for a fair die.

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Key Concepts

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

Discrete Random Variable
A discrete random variable is a type of variable that represents outcomes counted in integers. Think about rolling a die. Each face of the die represents one possible outcome, for instance, the numbers 1 to 6. For a discrete random variable, these specific outcomes can be enumerated: just like you can list each number on the die.

The probability associated with each of these outcomes can also be determined. In many scenarios, like our fair die example, each number is equally likely to occur, but in other situations, different probabilities might be assigned to various outcomes. What's important is that the variable can only take on particular values. It is not continuous, meaning it doesn’t include any numbers between the integers you've listed. Thus, you can have 3, 5, but not 3.5 or 5.2.
Probability Distribution Function
A probability distribution function, or PDF, assigns probabilities to all possible values of a discrete random variable. For a fair die, this is straightforward. Each side of the die is equally likely to show up, meaning it displays a uniform distribution over the outcomes 1 through 6.

In mathematical terms, the PDF for a fair die is given as follows:
  • For any given outcome x (where x is 1, 2, 3, 4, 5, or 6), the probability, denoted by \(p(x)\), is equal to \(\frac{1}{6}\).
This makes sense because there are six outcomes, all equally likely. The essence of the probability distribution function is to map how likely different results are for experiments; in this case, each face of the die is equally likely to appear.
Fair Die
A fair die is one that has an equal probability for every outcome. Imagine a perfectly balanced die, where every roll gives you any number from 1 to 6 with no bias.

This fairness is assumed in most theoretical frameworks, ensuring that the probability for each side landing face up is equal. With a six-sided die, this means each outcome is equally probable at \(\frac{1}{6}\). Such fair games are ideal models for learning about probability because they simplify calculations and make conceptual understanding easier.
  • Equal probability for each outcome: \(\frac{1}{6}\) for numbers 1 through 6.
  • Uniform distribution simplifies the analysis in probability theory.
Fair dice provide a clear, unbiased way to explore fundamental probability concepts, crucial for proper statistical modeling.
Expected Value
The expected value is a central concept in probability, representing the average or 'mean' of a random variable over many trials. If you roll a die many times, the expected result is the average outcome you'd anticipate.

For a fair die, the expected value can be calculated by multiplying each outcome by its probability and summing these products:
  • \( E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} \).
  • When simplified, it equals 3.5.
This value, 3.5, isn’t a possible outcome of one roll, but it’s the average across infinite rolls. It's a crucial concept, helping in predicting the behavior of systems modeled by random variables.

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Most popular questions from this chapter

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