k), E(X2k+) = 0, k = 0,1,.... Find the (2k)! 6. Let X be a r.v. such that E(X2k) = 5 mgf of X and also its ch.f. Then deduce the distribution of X.
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![**Problem Statement:**
Let \( X \) be a random variable (r.v.) such that:
\[ E(X^{2k}) = \frac{(2k)!}{k!}, \]
\[ E(X^{2k+1}) = 0, \quad k = 0, 1, \ldots \]
Find the moment generating function (mgf) of \( X \) and also its characteristic function (chf). Then deduce the distribution of \( X \).
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**Solution Outline:**
1. **Understand the Given Information:**
- The even moments of \( X \): \( E(X^{2k}) = \frac{(2k)!}{k!} \)
- The odd moments of \( X \): \( E(X^{2k+1}) = 0 \)
2. **Moment Generating Function (mgf):**
- The mgf \( M_X(t) \) is defined as \( E(e^{tX}) \).
- Use the given moments to express \( M_X(t) \).
3. **Characteristic Function (chf):**
- The chf \( \varphi_X(t) \) is defined as \( E(e^{itX}) \).
- Use the results from the mgf to find \( \varphi_X(t) \).
4. **Deduce the Distribution of \( X \):**
- Use the explicit form of \( M_X(t) \) to identify the distribution type of \( X \).
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Begin by noting the following:
For \( k = 0 \):
\[ E(X^0) = 1 \]
For \( k = 1 \):
\[ E(X^2) = 2! / 1! = 2 \]
For \( k = 2 \):
\[ E(X^4) = 4! / 2! = 24 / 2 = 12 \]
And so on...
By recognizing these moments correspond to those of a specific type of distribution, we further compute and identify the form of \( M_X(t) \) and \( \varphi_X(t) \). Finally, we identify that \( X \) is normally distributed with mean 0 and variance 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7fd60ee-f31e-44eb-bb20-a90745e03165%2Fa292bd18-30df-404c-85bc-c83caacfa810%2F3khdyan.jpeg&w=3840&q=75)
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