Solution Found!
This exercise demonstrates that, in general, the results
Chapter 3, Problem 169E(choose chapter or problem)
This exercise demonstrates that, in general, the results provided by Tchebysheff's theorem cannot be improved upon. Let be a random variable such that
\(p(-1)=\frac{1}{18}, p(0)=\frac{16}{18}, p(1)=\frac{1}{18}\)
Equation transcription:
Text transcription:
p(-1)=frac{1}{18}, p(0)=frac{16}{18}, p(1)=frac{1}{18}
Questions & Answers
QUESTION:
This exercise demonstrates that, in general, the results provided by Tchebysheff's theorem cannot be improved upon. Let be a random variable such that
\(p(-1)=\frac{1}{18}, p(0)=\frac{16}{18}, p(1)=\frac{1}{18}\)
Equation transcription:
Text transcription:
p(-1)=frac{1}{18}, p(0)=frac{16}{18}, p(1)=frac{1}{18}
ANSWER:Solution:
Step 1 of 4:
Let Y be a random variable such that
P(-1) = , P(0) =
, P(1)=
.
We have to,
-
Show that E(Y)=0 and V(Y) =
.
-
Calculate P(|Y-
|
) and we have to compare this probability with lower bound provided by Tchebysheff’s theorem.
-
Construct a probability distribution for a random variable X that will yield P(|X-
|
=
.
-
Find, if any k>1 is specified, how can a random variable W be constructed so that P(|W-
|
=
.