Solution Found!
Consider a random variable with a geometric distribution
Chapter 4, Problem 6E(choose chapter or problem)
Consider a random variable with a geometric distribution (Section 3.5); that is,
\(p(y)=q^{y-1} p\), \(y=1\), 2, 3,...., \(0<p<1\)
a. Show that has distribution function \(F(y)\) such that \(F(i)=1-q^{i}\), \(i=0\), 1, 2,... and that, in general,
\(F(y)=\left\{\begin{array}{ll}
0, & y<0 \\
1-q^{i}, & i \leq y<i+1, \quad \text { for } i=0,1,2, \ldots .
\end{array}\right.\)
b. Show that the preceding cumulative distribution function has the properties given in Theorem .
Equation Transcription:
Text Transcription:
p(y)=q^y-1 p
y=1
0<p<1
F(y)
F(i)=1-q^1
i=0
F(y)=0
y<0
F(y)=1-q^i
i</=y<i+1
i=0
Questions & Answers
QUESTION:
Consider a random variable with a geometric distribution (Section 3.5); that is,
\(p(y)=q^{y-1} p\), \(y=1\), 2, 3,...., \(0<p<1\)
a. Show that has distribution function \(F(y)\) such that \(F(i)=1-q^{i}\), \(i=0\), 1, 2,... and that, in general,
\(F(y)=\left\{\begin{array}{ll}
0, & y<0 \\
1-q^{i}, & i \leq y<i+1, \quad \text { for } i=0,1,2, \ldots .
\end{array}\right.\)
b. Show that the preceding cumulative distribution function has the properties given in Theorem .
Equation Transcription:
Text Transcription:
p(y)=q^y-1 p
y=1
0<p<1
F(y)
F(i)=1-q^1
i=0
F(y)=0
y<0
F(y)=1-q^i
i</=y<i+1
i=0
ANSWER:
Solution:
Step 1 of 2:
Let Y be the random variable with a Geometric distribution, that is
P(y) = p , y = 1, 2, 3,...., o < p < 1.
- The claim is to show that Y has distribution function F(y) such that
F(i) = 1 - , i = 0, 1, 2..
Let, F(i) = P( Y i )
This we can write as
F(i) = P( Y i )
= 1 - P(Y > i)
= 1 - P( 1st i trials are failures)
= 1 -
Hence, F(i) = 1 - .