Solution Found!
Suppose that Y is a continuous random variable with
Chapter 4, Problem 34E(choose chapter or problem)
Suppose that is a continuous random variable with density \(f(y)\) that is positive only if \(y \geq 0\). If \(F(y)\) is the distribution function, show that
\(E(Y)=\int_{0}^{\infty} y f(y)\) \(d y=\int_{0}^{\infty}[1-F(y)] d y\).
[Hint: If \(y>0\), \(y=\int_{0}^{y} d t\), and \(E(Y)=\int_{0}^{\infty} y f(y)\) \(d y=\int_{0}^{\infty}\left\{\int_{0}^{y} d t\right\} f(y) d y\). Exchange the
order of integration to obtain the desired result.\(]^{4}\)
Equation Transcription:
Text Transcription:
f(y)
y>/=0
F(y)
E(Y)=integral_0^infinity yf(y)
dy=integral_0^infinity [1=F(y)]dy
y>0
y=integral_0^y dt
E(Y)=integral_0^infinity yf(y)
dy=integral_0^infinity{integral_0^y dt}f(y)dy
Questions & Answers
QUESTION:
Suppose that is a continuous random variable with density \(f(y)\) that is positive only if \(y \geq 0\). If \(F(y)\) is the distribution function, show that
\(E(Y)=\int_{0}^{\infty} y f(y)\) \(d y=\int_{0}^{\infty}[1-F(y)] d y\).
[Hint: If \(y>0\), \(y=\int_{0}^{y} d t\), and \(E(Y)=\int_{0}^{\infty} y f(y)\) \(d y=\int_{0}^{\infty}\left\{\int_{0}^{y} d t\right\} f(y) d y\). Exchange the
order of integration to obtain the desired result.\(]^{4}\)
Equation Transcription:
Text Transcription:
f(y)
y>/=0
F(y)
E(Y)=integral_0^infinity yf(y)
dy=integral_0^infinity [1=F(y)]dy
y>0
y=integral_0^y dt
E(Y)=integral_0^infinity yf(y)
dy=integral_0^infinity{integral_0^y dt}f(y)dy
ANSWER:
Answer:
Step 1 of 1:
Suppose that with density
that is positive only if
If
a density function, show that
The expected value of a is
Provided that the integral exists.
Since with density
that is positive only if