How to understand conditional distribution of exponential where it is conditioned on being less or greater...
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From SOA sample #270
The lifetime of a machine part is exponentially distributed with a mean of five years.
Calculate the mean lifetime of the part, given that it survives less than ten years.
To find the conditional distribution, I must define $$P(X|X<10)={P(X=xcap X<10)over P(X<10)}$$
The denominator is $1-e^{-2}$. Regarding the numerator, the SOA solution says it is simply the exponential distribution of mean $5$, ie. $frac1{5}e^{-frac x{5}}$. This confuses me, because how do we take into account the second part of the numerator, the condition of $X<10$? Can we simply define this expression $frac1{5}e^{-frac x{5}}$, as only applying when $0<x<10$, without modifying the expression in any way?
The answer here seems to do it a different way.
probability statistics
asked Dec 3 '18 at 4:47
agbltagblt
17914
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add a comment |
$begingroup$
From SOA sample #270
The lifetime of a machine part is exponentially distributed with a mean of five years.
Calculate the mean lifetime of the part, given that it survives less than ten years.
To find the conditional distribution, I must define $$P(X|X<10)={P(X=xcap X<10)over P(X<10)}$$
The denominator is $1-e^{-2}$. Regarding the numerator, the SOA solution says it is simply the exponential distribution of mean $5$, ie. $frac1{5}e^{-frac x{5}}$. This confuses me, because how do we take into account the second part of the numerator, the condition of $X<10$? Can we simply define this expression $frac1{5}e^{-frac x{5}}$, as only applying when $0<x<10$, without modifying the expression in any way?
The answer here seems to do it a different way.
probability statistics
asked Dec 3 '18 at 4:47
agbltagblt
17914
$endgroup$
$begingroup$
What is "complicated" in the answer over there?
$endgroup$
– Did
Dec 3 '18 at 7:02
$begingroup$
After looking at it again, it's not.
$endgroup$
– agblt
Dec 3 '18 at 11:32
add a comment |
$begingroup$
From SOA sample #270
The lifetime of a machine part is exponentially distributed with a mean of five years.
Calculate the mean lifetime of the part, given that it survives less than ten years.
To find the conditional distribution, I must define $$P(X|X<10)={P(X=xcap X<10)over P(X<10)}$$
The denominator is $1-e^{-2}$. Regarding the numerator, the SOA solution says it is simply the exponential distribution of mean $5$, ie. $frac1{5}e^{-frac x{5}}$. This confuses me, because how do we take into account the second part of the numerator, the condition of $X<10$? Can we simply define this expression $frac1{5}e^{-frac x{5}}$, as only applying when $0<x<10$, without modifying the expression in any way?
The answer here seems to do it a different way.
probability statistics
asked Dec 3 '18 at 4:47
agbltagblt
17914
$endgroup$
From SOA sample #270
The lifetime of a machine part is exponentially distributed with a mean of five years.
Calculate the mean lifetime of the part, given that it survives less than ten years.
To find the conditional distribution, I must define $$P(X|X<10)={P(X=xcap X<10)over P(X<10)}$$
The denominator is $1-e^{-2}$. Regarding the numerator, the SOA solution says it is simply the exponential distribution of mean $5$, ie. $frac1{5}e^{-frac x{5}}$. This confuses me, because how do we take into account the second part of the numerator, the condition of $X<10$? Can we simply define this expression $frac1{5}e^{-frac x{5}}$, as only applying when $0<x<10$, without modifying the expression in any way?
The answer here seems to do it a different way.
probability statistics
probability statistics
asked Dec 3 '18 at 4:47
agbltagblt
17914
asked Dec 3 '18 at 4:47
agbltagblt
17914
edited Dec 3 '18 at 11:27
agblt
asked Dec 3 '18 at 4:47
agbltagblt
17914
asked Dec 3 '18 at 4:47
agbltagblt
17914
asked Dec 3 '18 at 4:47
agbltagblt
17914
17914
$begingroup$
What is "complicated" in the answer over there?
$endgroup$
– Did
Dec 3 '18 at 7:02
$begingroup$
After looking at it again, it's not.
$endgroup$
– agblt
Dec 3 '18 at 11:32
add a comment |
$begingroup$
What is "complicated" in the answer over there?
$endgroup$
– Did
Dec 3 '18 at 7:02
$begingroup$
After looking at it again, it's not.
$endgroup$
– agblt
Dec 3 '18 at 11:32
$begingroup$
What is "complicated" in the answer over there?
$endgroup$
– Did
Dec 3 '18 at 7:02
$begingroup$
What is "complicated" in the answer over there?
$endgroup$
– Did
Dec 3 '18 at 7:02
$begingroup$
After looking at it again, it's not.
$endgroup$
– agblt
Dec 3 '18 at 11:32
$begingroup$
After looking at it again, it's not.
$endgroup$
– agblt
Dec 3 '18 at 11:32
add a comment |
1 Answer
1
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votes
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You need to consider the probability distribution $color{blue}{restricted}$ to $[0,10)$ but besides this you need to $color{blue}{normalize}$ it by $P([0,10)) = 1-frac{1}{e^2}$.
So, the probability density of the conditioned random variable is:
$$f_{X|X<10}(x)= begin{cases}
0 & x geq 10 \
frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} e^{-frac{x}{5}} & 0 color{blue}{leq x <10} \
0 & x < 10
end{cases}$$
Now,
$$E(X|X<10) = int_0^{infty}xf_{X|X<10}(x), dx = frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} int_0^{color{blue}{10}} x e^{-frac{x}{5}} , dx$$
answered Dec 3 '18 at 5:06
trancelocationtrancelocation
10.5k1722
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need to consider the probability distribution $color{blue}{restricted}$ to $[0,10)$ but besides this you need to $color{blue}{normalize}$ it by $P([0,10)) = 1-frac{1}{e^2}$.
So, the probability density of the conditioned random variable is:
$$f_{X|X<10}(x)= begin{cases}
0 & x geq 10 \
frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} e^{-frac{x}{5}} & 0 color{blue}{leq x <10} \
0 & x < 10
end{cases}$$
Now,
$$E(X|X<10) = int_0^{infty}xf_{X|X<10}(x), dx = frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} int_0^{color{blue}{10}} x e^{-frac{x}{5}} , dx$$
answered Dec 3 '18 at 5:06
trancelocationtrancelocation
10.5k1722
$endgroup$
add a comment |
$begingroup$
You need to consider the probability distribution $color{blue}{restricted}$ to $[0,10)$ but besides this you need to $color{blue}{normalize}$ it by $P([0,10)) = 1-frac{1}{e^2}$.
So, the probability density of the conditioned random variable is:
$$f_{X|X<10}(x)= begin{cases}
0 & x geq 10 \
frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} e^{-frac{x}{5}} & 0 color{blue}{leq x <10} \
0 & x < 10
end{cases}$$
Now,
$$E(X|X<10) = int_0^{infty}xf_{X|X<10}(x), dx = frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} int_0^{color{blue}{10}} x e^{-frac{x}{5}} , dx$$
answered Dec 3 '18 at 5:06
trancelocationtrancelocation
10.5k1722
$endgroup$
add a comment |
$begingroup$
You need to consider the probability distribution $color{blue}{restricted}$ to $[0,10)$ but besides this you need to $color{blue}{normalize}$ it by $P([0,10)) = 1-frac{1}{e^2}$.
So, the probability density of the conditioned random variable is:
$$f_{X|X<10}(x)= begin{cases}
0 & x geq 10 \
frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} e^{-frac{x}{5}} & 0 color{blue}{leq x <10} \
0 & x < 10
end{cases}$$
Now,
$$E(X|X<10) = int_0^{infty}xf_{X|X<10}(x), dx = frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} int_0^{color{blue}{10}} x e^{-frac{x}{5}} , dx$$
answered Dec 3 '18 at 5:06
trancelocationtrancelocation
10.5k1722
$endgroup$
You need to consider the probability distribution $color{blue}{restricted}$ to $[0,10)$ but besides this you need to $color{blue}{normalize}$ it by $P([0,10)) = 1-frac{1}{e^2}$.
So, the probability density of the conditioned random variable is:
$$f_{X|X<10}(x)= begin{cases}
0 & x geq 10 \
frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} e^{-frac{x}{5}} & 0 color{blue}{leq x <10} \
0 & x < 10
end{cases}$$
Now,
$$E(X|X<10) = int_0^{infty}xf_{X|X<10}(x), dx = frac{1}{5left(color{blue}{1-frac{1}{e^2}}right)} int_0^{color{blue}{10}} x e^{-frac{x}{5}} , dx$$
answered Dec 3 '18 at 5:06
trancelocationtrancelocation
10.5k1722
edited Dec 3 '18 at 5:22
answered Dec 3 '18 at 5:06
trancelocationtrancelocation
10.5k1722
answered Dec 3 '18 at 5:06
trancelocationtrancelocation
10.5k1722
answered Dec 3 '18 at 5:06
trancelocationtrancelocation
10.5k1722
10.5k1722
add a comment |
add a comment |
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$begingroup$
What is "complicated" in the answer over there?
$endgroup$
– Did
Dec 3 '18 at 7:02
$begingroup$
After looking at it again, it's not.
$endgroup$
– agblt
Dec 3 '18 at 11:32