Unit 4: Moments - Subjective Questions
ECAP790 • Practice Questions with Detailed Answers
20 questions
Define the th raw moment and the th central moment of a random variable . Distinguish between them and state their significance.
Raw moment: The th raw moment of about the origin is
Central moment: The th central moment of about its mean is
Distinction:
- Raw moments are measured about the origin, whereas central moments are measured about the mean.
- The first raw moment is the mean: .
- The first central moment is always zero: .
- The second central moment is the variance: .
- Higher central moments describe the shape of a distribution. The third relates to skewness, while the fourth relates to kurtosis.
Derive the relationships between the first four central moments and the raw moments of a random variable.
Let . Using
and expanding by the binomial theorem gives:
- First central moment:
- Second central moment:
- Third central moment:
- Fourth central moment:
In general,
where .
State and prove Chebyshev's inequality for a random variable having finite mean and variance.
Chebyshev's inequality: If has mean and finite variance , then for every ,
Equivalently, for ,
Proof: Define the nonnegative random variable . By Markov's inequality,
Since and
we obtain
Thus, at least of the probability lies within standard deviations of the mean.
A random variable has mean and variance . Use Chebyshev's inequality to find a lower bound for and an upper bound for .
The mean and standard deviation are
For the first probability,
Here . Chebyshev's inequality gives
Therefore,
For the second probability, , so
These are distribution-free bounds; no assumption about the form of the distribution is required.
Explain joint moments and mixed central moments of two random variables. Show how covariance is expressed as a joint central moment.
For random variables and , the joint raw moment of order is
The corresponding mixed central moment is
Important special cases include:
- and .
- .
- .
- The mixed central moment of order is
In terms of raw moments,
Thus, covariance measures the second-order joint variation of and .
If and are independent random variables, prove that their product moments factorize. Does zero covariance imply independence?
If and are independent and the required moments exist, their joint distribution factorizes. Therefore,
For continuous variables, for example,
Taking gives
and hence . Thus, independence implies zero covariance when second moments exist.
The converse is generally false. For example, let be uniformly distributed on and set . Symmetry gives and , so
However, is completely determined by , so and are not independent.
Derive the first and second raw moments of the sum in terms of the moments of and .
By linearity of expectation,
Thus, the first raw moment is
For the second raw moment,
Taking expectations,
Therefore,
where . If and are independent, then , and
Derive the mean and variance of a linear combination . State the result when the variables are independent.
By linearity of expectation,
For the variance,
Expanding the square gives
Equivalently, if is the coefficient vector and is the covariance matrix,
If the variables are independent, all pairwise covariance terms are zero, so
For , this reduces to the variance formula for a sum.
Define the moment generating function of a random variable. Explain how it generates raw moments and state the relevant existence condition.
The moment generating function (MGF) of is
for those values of for which the expectation is finite.
Expanding the exponential,
Under conditions that justify interchanging expectation and summation,
Consequently, the th raw moment is obtained by differentiation:
Important points are:
- .
- An MGF is normally required to be finite on an open interval containing .
- If it exists on such an interval, it uniquely determines the distribution.
- The existence of all ordinary moments alone does not necessarily imply that the MGF exists near .
Prove the transformation property of the moment generating function for and use it to obtain the mean and variance of .
For ,
Separating the constant term gives
The mean can be found directly or by differentiating at zero:
Similarly,
Therefore,
Thus, adding shifts the mean but does not affect variance, while multiplication by multiplies the variance by .
Show that the MGF of a sum of independent random variables is the product of their individual MGFs. Explain why independence is essential.
Let
Then
Because are independent, the transformed variables are also independent. Hence,
Independence is essential because it permits the expectation of the product to be written as the product of expectations. Without independence, the joint dependence structure affects , and the product formula need not hold.
If the variables are independent and identically distributed with MGF , then
Use the moment generating function to derive the first four raw moments of a normal random variable .
The MGF of is
Raw moments are obtained from . Differentiation gives:
and
Therefore, the first four raw moments are
The corresponding third central moment is , and the fourth central moment is .
Define the cumulant generating function and derive the first four cumulants in terms of central moments.
The cumulant generating function (CGF) is the logarithm of the MGF:
Its series expansion is
where
is the th cumulant. In terms of the mean and central moments:
- First cumulant:
- Second cumulant:
- Third cumulant:
- Fourth cumulant:
Thus, the first cumulant is the mean, the second is the variance, and the higher cumulants measure distributional characteristics beyond location and dispersion.
Establish the additivity property of cumulants for independent random variables and state the effect of a linear transformation on cumulants.
Let and be independent. Since
taking logarithms gives
Differentiating times at yields
More generally, cumulants of a sum of independent variables are the sums of the corresponding cumulants.
For ,
Therefore,
and, for ,
The shift affects only the first cumulant, while scaling affects the th cumulant by the factor .
Find the cumulant generating function and the first four cumulants of a Poisson random variable with parameter .
For , the MGF is
Therefore, the cumulant generating function is
Differentiating,
for every integer . Evaluating at gives
Hence, the first four cumulants are
Thus, the Poisson distribution has mean and variance both equal to . Its third central moment is , while its fourth central moment is
Define the joint moment generating function of two random variables and explain how it is used to obtain means, variances, covariance, and mixed moments.
The joint MGF of and is
provided it is finite in a neighborhood of . Mixed raw moments are obtained through partial differentiation:
In particular,
and
Therefore,
If the joint MGF factorizes as near the origin, then and are independent.
Explain why the existence of all moments does not necessarily guarantee the existence of an MGF. Illustrate using the lognormal distribution.
A random variable may possess finite moments of every positive integer order while its MGF is infinite for every .
Let , where . Then has a lognormal distribution. Its th raw moment is
which is finite for every positive integer .
However, for ,
diverges because grows too rapidly in the upper tail. Hence, the MGF is not finite on any open interval around .
This demonstrates that:
- Finite moments of all orders do not guarantee MGF existence near zero.
- The formal moment series need not converge to an MGF.
- When the MGF does exist in a neighborhood of zero, it uniquely determines the distribution.
Define the coefficients of skewness and kurtosis in terms of central moments. Interpret their values and relate kurtosis to the fourth cumulant.
Let , , and denote the second, third, and fourth central moments.
The coefficient of skewness is
Its interpretation is:
- : symmetric in terms of the third central moment.
- : positive or right skewness.
- : negative or left skewness.
The coefficient of kurtosis is
and excess kurtosis is
A normal distribution has and . Since
standardized excess kurtosis can also be written as
For random variables , , and , derive the third raw moment of their sum . Simplify the expression when the variables are independent.
Expanding the cube,
Therefore,
Taking expectations gives
If , , and are mutually independent, the mixed moments factorize. Thus,
with analogous expressions for the other terms. Substitution produces the third raw moment entirely in terms of the individual raw moments.
Let be independent and identically distributed random variables with mean , variance , and finite third and fourth cumulants. Determine the first four cumulants of their standardized sum and explain their limiting behavior.
Define the standardized sum
Let the cumulants of each be , , , and . Additivity of cumulants gives cumulant for the unstandardized sum.
Centering and scaling then yield:
and
As , the third and fourth cumulants approach zero, while the first two approach and . These are the cumulants of the standard normal distribution, illustrating the moment-cumulant basis of normal convergence under suitable conditions.
Define the th raw moment and the th central moment of a random variable . Distinguish between them and state their significance.
Raw moment: The th raw moment of about the origin is
Central moment: The th central moment of about its mean is
Distinction:
- Raw moments are measured about the origin, whereas central moments are measured about the mean.
- The first raw moment is the mean: .
- The first central moment is always zero: .
- The second central moment is the variance: .
- Higher central moments describe the shape of a distribution. The third relates to skewness, while the fourth relates to kurtosis.
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