Solution Found!
Suppose that Y is a discrete random variable with mean ?
Chapter 3, Problem 30E(choose chapter or problem)
Suppose that is a discrete random variable with mean
and variance \(\sigma^{2}\) and let \(X=Y+1\).
a Do you expect the mean of to be larger than, smaller than, or equal to \(\mu=E(Y)\)? Why?
b Use Theorems and
to express \(E(X)=E(Y+1)\) in terms of \(\mu=E(Y)\). Does this result agree with your answer to part (a)?
c Recalling that the variance is a measure of spread or dispersion, do you expect the variance of to be larger than, smaller than, or equal to \(\sigma^{2}=V(Y)\)? Why?
3. Exercises preceded by an asterisk are optional.
d Use Definition and the result in part (b) to show that
\(V(X)=E\left\{[X-E(X)]^{2}\right\}=E\left[(Y-\mu)^{2}\right]=\sigma^{2}\);
that is, \(X=Y+1\) and have equal variances.
Equation Transcription:
Text Transcription:
sigma^2
X=Y+1
mu=E(Y)
E(X)=E(Y+1)
mu=E(Y)
sigma^2=V(Y)
V(X)=E{[X-E(X)]^2}=E[(Y-)^2]=sigma^2
X=Y+1
Questions & Answers
QUESTION:
Suppose that is a discrete random variable with mean
and variance \(\sigma^{2}\) and let \(X=Y+1\).
a Do you expect the mean of to be larger than, smaller than, or equal to \(\mu=E(Y)\)? Why?
b Use Theorems and
to express \(E(X)=E(Y+1)\) in terms of \(\mu=E(Y)\). Does this result agree with your answer to part (a)?
c Recalling that the variance is a measure of spread or dispersion, do you expect the variance of to be larger than, smaller than, or equal to \(\sigma^{2}=V(Y)\)? Why?
3. Exercises preceded by an asterisk are optional.
d Use Definition and the result in part (b) to show that
\(V(X)=E\left\{[X-E(X)]^{2}\right\}=E\left[(Y-\mu)^{2}\right]=\sigma^{2}\);
that is, \(X=Y+1\) and have equal variances.
Equation Transcription:
Text Transcription:
sigma^2
X=Y+1
mu=E(Y)
E(X)=E(Y+1)
mu=E(Y)
sigma^2=V(Y)
V(X)=E{[X-E(X)]^2}=E[(Y-)^2]=sigma^2
X=Y+1
ANSWER:
Solution:
Step 1 of 3:
Given, let Y is a discrete random variable with mean and variance
,
X = Y + 1
-
The claim is to suggest that the mean of the X to be larger than, smaller than, or equal to
= E(Y).
We have , X = Y + 1
Take expectation
E(X) = E(Y + 1)
= E(Y) + 1
We know that, E(Y) =
Therefore, E(X) = + 1, which is greater than
.
Hence, the mean of the X to be larger than = E(Y).