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Ch 4.12 - 233SE
Chapter 4, Problem 233SE(choose chapter or problem)
The two-parameter exponential distribution uses a different range for the random variable
\(\text { X }\), namely, \(0 \leq y \leq x\) for a constant γ (and this equals the usual exponential distribution in the special case that
\(y=0\)). The probability density function for \(\text { X }\)is f \((x)=\lambda\) \(\exp [-\lambda(x-y)]\) for \(0 \leq y \leq x\) and
\(0<\lambda\) . Determine the following in terms of the parameters \(\lambda \text { and } \mathrm{Y}\):
(a) Mean and variance of \(\text { X }\).
(b) \(P(X<y+1 / \lambda)\)
Equation Transcription:
Text Transcription:
X
0 \leq y \leq x
y
y=0
X
(x)=\lambda
Exp[- \lambda (x-y)]
0 \leq y \leq x
0<\lambda
\lambda and y
X
P(X<y+1/\lambda)
Questions & Answers
QUESTION:
The two-parameter exponential distribution uses a different range for the random variable
\(\text { X }\), namely, \(0 \leq y \leq x\) for a constant γ (and this equals the usual exponential distribution in the special case that
\(y=0\)). The probability density function for \(\text { X }\)is f \((x)=\lambda\) \(\exp [-\lambda(x-y)]\) for \(0 \leq y \leq x\) and
\(0<\lambda\) . Determine the following in terms of the parameters \(\lambda \text { and } \mathrm{Y}\):
(a) Mean and variance of \(\text { X }\).
(b) \(P(X<y+1 / \lambda)\)
Equation Transcription:
Text Transcription:
X
0 \leq y \leq x
y
y=0
X
(x)=\lambda
Exp[- \lambda (x-y)]
0 \leq y \leq x
0<\lambda
\lambda and y
X
P(X<y+1/\lambda)
ANSWER:
Solution:
Step 1 of 4:
The probability density function for X is,
f(x) = , for 0
and 0 <
.
- We have to find the mean and variance of X.
-
We have to find P(X<
+
).