PDF of an exponential distribution with varying paramter, lambda
Suppose that the lifetime of a device is exponential with rate λ, but suppose
also that the value of λ is not fixed but is itself a random variable that is
uniform in the range [a, b) with 0 < a.
Can the pdf of the lifetime of the device be written as:
$ lambda = begin{cases} 0, x < a\ dfrac {1} {b-a} , aleq x < b\ 0, xgeq b end{cases} $
probability uniform-distribution exponential-distribution
asked Nov 30 '18 at 1:11
BasileusBasileus
113
add a comment |
Suppose that the lifetime of a device is exponential with rate λ, but suppose
also that the value of λ is not fixed but is itself a random variable that is
uniform in the range [a, b) with 0 < a.
Can the pdf of the lifetime of the device be written as:
$ lambda = begin{cases} 0, x < a\ dfrac {1} {b-a} , aleq x < b\ 0, xgeq b end{cases} $
probability uniform-distribution exponential-distribution
asked Nov 30 '18 at 1:11
BasileusBasileus
113
1
The PDF of $lambda$, or the PDF of the lifetime of the device?
– Clarinetist
Nov 30 '18 at 1:12
If $lambda$ is Uniform on $[a,b)$, then the PDF would be 0 for $xgeq b$.
– gd1035
Nov 30 '18 at 1:18
What you wrote is the PDF of the random variable $lambda$, not the PDF of the device lifetime.
– BlackMath
Nov 30 '18 at 6:06
add a comment |
Suppose that the lifetime of a device is exponential with rate λ, but suppose
also that the value of λ is not fixed but is itself a random variable that is
uniform in the range [a, b) with 0 < a.
Can the pdf of the lifetime of the device be written as:
$ lambda = begin{cases} 0, x < a\ dfrac {1} {b-a} , aleq x < b\ 0, xgeq b end{cases} $
probability uniform-distribution exponential-distribution
asked Nov 30 '18 at 1:11
BasileusBasileus
113
Suppose that the lifetime of a device is exponential with rate λ, but suppose
also that the value of λ is not fixed but is itself a random variable that is
uniform in the range [a, b) with 0 < a.
Can the pdf of the lifetime of the device be written as:
$ lambda = begin{cases} 0, x < a\ dfrac {1} {b-a} , aleq x < b\ 0, xgeq b end{cases} $
probability uniform-distribution exponential-distribution
probability uniform-distribution exponential-distribution
asked Nov 30 '18 at 1:11
BasileusBasileus
113
asked Nov 30 '18 at 1:11
BasileusBasileus
113
edited Nov 30 '18 at 1:25
Basileus
asked Nov 30 '18 at 1:11
BasileusBasileus
113
asked Nov 30 '18 at 1:11
BasileusBasileus
113
asked Nov 30 '18 at 1:11
BasileusBasileus
113
113
1
The PDF of $lambda$, or the PDF of the lifetime of the device?
– Clarinetist
Nov 30 '18 at 1:12
If $lambda$ is Uniform on $[a,b)$, then the PDF would be 0 for $xgeq b$.
– gd1035
Nov 30 '18 at 1:18
What you wrote is the PDF of the random variable $lambda$, not the PDF of the device lifetime.
– BlackMath
Nov 30 '18 at 6:06
add a comment |
1
The PDF of $lambda$, or the PDF of the lifetime of the device?
– Clarinetist
Nov 30 '18 at 1:12
If $lambda$ is Uniform on $[a,b)$, then the PDF would be 0 for $xgeq b$.
– gd1035
Nov 30 '18 at 1:18
What you wrote is the PDF of the random variable $lambda$, not the PDF of the device lifetime.
– BlackMath
Nov 30 '18 at 6:06
1
1
The PDF of $lambda$, or the PDF of the lifetime of the device?
– Clarinetist
Nov 30 '18 at 1:12
The PDF of $lambda$, or the PDF of the lifetime of the device?
– Clarinetist
Nov 30 '18 at 1:12
If $lambda$ is Uniform on $[a,b)$, then the PDF would be 0 for $xgeq b$.
– gd1035
Nov 30 '18 at 1:18
If $lambda$ is Uniform on $[a,b)$, then the PDF would be 0 for $xgeq b$.
– gd1035
Nov 30 '18 at 1:18
What you wrote is the PDF of the random variable $lambda$, not the PDF of the device lifetime.
– BlackMath
Nov 30 '18 at 6:06
What you wrote is the PDF of the random variable $lambda$, not the PDF of the device lifetime.
– BlackMath
Nov 30 '18 at 6:06
add a comment |
1 Answer
1
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oldest
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Let $T|lambda sim text{Exp}(lambda)$ be the lifetime of the device and let $lambda sim text{U}[a,b)$ be the parameter. Then for all $t geqslant 0$ you have:
$$begin{equation} begin{aligned}
f_T(t)
&= int limits_a^b p(T=t|lambda) pi(lambda) dlambda \[6pt]
&= frac{1}{a-b} int limits_a^b exp(-lambda t) dlambda \[6pt]
&= frac{1}{a-b} Bigg[ - frac{1}{t} cdot exp(-lambda t) Bigg]_{lambda=a}^{lambda=b} \[6pt]
&= frac{1}{a-b} Bigg[ frac{exp(-at) - exp(-bt)}{t} Bigg] \[6pt]
&= frac{e^{-at} - e^{-bt}}{a-b} cdot frac{1}{t}. \[6pt]
end{aligned} end{equation}$$
answered Nov 30 '18 at 5:45
BenBen
92611
So, this has to take into account the law of total probability, thanks for that.
– Basileus
Nov 30 '18 at 13:09
Yes, the first step is an application of that law.
– Ben
Nov 30 '18 at 22:51
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
active
oldest
votes
Let $T|lambda sim text{Exp}(lambda)$ be the lifetime of the device and let $lambda sim text{U}[a,b)$ be the parameter. Then for all $t geqslant 0$ you have:
$$begin{equation} begin{aligned}
f_T(t)
&= int limits_a^b p(T=t|lambda) pi(lambda) dlambda \[6pt]
&= frac{1}{a-b} int limits_a^b exp(-lambda t) dlambda \[6pt]
&= frac{1}{a-b} Bigg[ - frac{1}{t} cdot exp(-lambda t) Bigg]_{lambda=a}^{lambda=b} \[6pt]
&= frac{1}{a-b} Bigg[ frac{exp(-at) - exp(-bt)}{t} Bigg] \[6pt]
&= frac{e^{-at} - e^{-bt}}{a-b} cdot frac{1}{t}. \[6pt]
end{aligned} end{equation}$$
answered Nov 30 '18 at 5:45
BenBen
92611
So, this has to take into account the law of total probability, thanks for that.
– Basileus
Nov 30 '18 at 13:09
Yes, the first step is an application of that law.
– Ben
Nov 30 '18 at 22:51
add a comment |
Let $T|lambda sim text{Exp}(lambda)$ be the lifetime of the device and let $lambda sim text{U}[a,b)$ be the parameter. Then for all $t geqslant 0$ you have:
$$begin{equation} begin{aligned}
f_T(t)
&= int limits_a^b p(T=t|lambda) pi(lambda) dlambda \[6pt]
&= frac{1}{a-b} int limits_a^b exp(-lambda t) dlambda \[6pt]
&= frac{1}{a-b} Bigg[ - frac{1}{t} cdot exp(-lambda t) Bigg]_{lambda=a}^{lambda=b} \[6pt]
&= frac{1}{a-b} Bigg[ frac{exp(-at) - exp(-bt)}{t} Bigg] \[6pt]
&= frac{e^{-at} - e^{-bt}}{a-b} cdot frac{1}{t}. \[6pt]
end{aligned} end{equation}$$
answered Nov 30 '18 at 5:45
BenBen
92611
So, this has to take into account the law of total probability, thanks for that.
– Basileus
Nov 30 '18 at 13:09
Yes, the first step is an application of that law.
– Ben
Nov 30 '18 at 22:51
add a comment |
Let $T|lambda sim text{Exp}(lambda)$ be the lifetime of the device and let $lambda sim text{U}[a,b)$ be the parameter. Then for all $t geqslant 0$ you have:
$$begin{equation} begin{aligned}
f_T(t)
&= int limits_a^b p(T=t|lambda) pi(lambda) dlambda \[6pt]
&= frac{1}{a-b} int limits_a^b exp(-lambda t) dlambda \[6pt]
&= frac{1}{a-b} Bigg[ - frac{1}{t} cdot exp(-lambda t) Bigg]_{lambda=a}^{lambda=b} \[6pt]
&= frac{1}{a-b} Bigg[ frac{exp(-at) - exp(-bt)}{t} Bigg] \[6pt]
&= frac{e^{-at} - e^{-bt}}{a-b} cdot frac{1}{t}. \[6pt]
end{aligned} end{equation}$$
answered Nov 30 '18 at 5:45
BenBen
92611
Let $T|lambda sim text{Exp}(lambda)$ be the lifetime of the device and let $lambda sim text{U}[a,b)$ be the parameter. Then for all $t geqslant 0$ you have:
$$begin{equation} begin{aligned}
f_T(t)
&= int limits_a^b p(T=t|lambda) pi(lambda) dlambda \[6pt]
&= frac{1}{a-b} int limits_a^b exp(-lambda t) dlambda \[6pt]
&= frac{1}{a-b} Bigg[ - frac{1}{t} cdot exp(-lambda t) Bigg]_{lambda=a}^{lambda=b} \[6pt]
&= frac{1}{a-b} Bigg[ frac{exp(-at) - exp(-bt)}{t} Bigg] \[6pt]
&= frac{e^{-at} - e^{-bt}}{a-b} cdot frac{1}{t}. \[6pt]
end{aligned} end{equation}$$
answered Nov 30 '18 at 5:45
BenBen
92611
answered Nov 30 '18 at 5:45
BenBen
92611
answered Nov 30 '18 at 5:45
BenBen
92611
answered Nov 30 '18 at 5:45
BenBen
92611
92611
So, this has to take into account the law of total probability, thanks for that.
– Basileus
Nov 30 '18 at 13:09
Yes, the first step is an application of that law.
– Ben
Nov 30 '18 at 22:51
add a comment |
So, this has to take into account the law of total probability, thanks for that.
– Basileus
Nov 30 '18 at 13:09
Yes, the first step is an application of that law.
– Ben
Nov 30 '18 at 22:51
So, this has to take into account the law of total probability, thanks for that.
– Basileus
Nov 30 '18 at 13:09
So, this has to take into account the law of total probability, thanks for that.
– Basileus
Nov 30 '18 at 13:09
Yes, the first step is an application of that law.
– Ben
Nov 30 '18 at 22:51
Yes, the first step is an application of that law.
– Ben
Nov 30 '18 at 22:51
add a comment |
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1
The PDF of $lambda$, or the PDF of the lifetime of the device?
– Clarinetist
Nov 30 '18 at 1:12
If $lambda$ is Uniform on $[a,b)$, then the PDF would be 0 for $xgeq b$.
– gd1035
Nov 30 '18 at 1:18
What you wrote is the PDF of the random variable $lambda$, not the PDF of the device lifetime.
– BlackMath
Nov 30 '18 at 6:06