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Exponential conditional probability
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There are two types of claims that are made to an insurance company. Let $N_i(t)$ denote the number of type $i$ claims made by time $t$, and suppose that $N_1(t), t ge 0$ and $N_2(t), t ge 0$ are independent Poisson processes with rates $λ_1 = 10$ and $λ_2 = 1$.
The amounts of successive type 1 claims are independent exponential
random variables with mean 1000 dollars whereas the amounts from type 2 claims are independent exponential random variables with mean 5000 dollars. A claim for 4000 dollars has just been received; what is the probability it is a type 1 claim?
Here's my approach to the problem:
beginalign
& P(textclaim = texttype 1mid 4000) \[10pt]
= & fracP(4000mid textclaim = texttype 1)(textclaim = texttype 1)P(4000mid textclaim = texttype 1)P(textclaim = texttype 1) + P(4000midtextclaim = texttype 2)P(textclaim = texttype 2)
endalign
by Bayes' formula.
My question is, how do I calculate $P(4000mid textclaim = texttype 1)$ and $P(4000mid textclaim = texttype 2)$? The dollar value of claims is exponentially distributed, and since the distribution is continuous I can't fix a value to the pdf, I need a range. Any tips on how to calculate these two values?
probability statistics bayes-theorem exponential-distribution
edited Jul 25 at 3:12
Michael Hardy
204k23186461
asked Jul 25 at 3:02
Andrew Louis
656
add a comment |Â
up vote
3
down vote
favorite
There are two types of claims that are made to an insurance company. Let $N_i(t)$ denote the number of type $i$ claims made by time $t$, and suppose that $N_1(t), t ge 0$ and $N_2(t), t ge 0$ are independent Poisson processes with rates $λ_1 = 10$ and $λ_2 = 1$.
The amounts of successive type 1 claims are independent exponential
random variables with mean 1000 dollars whereas the amounts from type 2 claims are independent exponential random variables with mean 5000 dollars. A claim for 4000 dollars has just been received; what is the probability it is a type 1 claim?
Here's my approach to the problem:
beginalign
& P(textclaim = texttype 1mid 4000) \[10pt]
= & fracP(4000mid textclaim = texttype 1)(textclaim = texttype 1)P(4000mid textclaim = texttype 1)P(textclaim = texttype 1) + P(4000midtextclaim = texttype 2)P(textclaim = texttype 2)
endalign
by Bayes' formula.
My question is, how do I calculate $P(4000mid textclaim = texttype 1)$ and $P(4000mid textclaim = texttype 2)$? The dollar value of claims is exponentially distributed, and since the distribution is continuous I can't fix a value to the pdf, I need a range. Any tips on how to calculate these two values?
probability statistics bayes-theorem exponential-distribution
edited Jul 25 at 3:12
Michael Hardy
204k23186461
asked Jul 25 at 3:02
Andrew Louis
656
It is pretty common to encounter continuous distributions in Bayes distributions. Just take the PDF value. It will do because the range cancels out from the numerator and denominator.
– kasa
Jul 25 at 3:12
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
There are two types of claims that are made to an insurance company. Let $N_i(t)$ denote the number of type $i$ claims made by time $t$, and suppose that $N_1(t), t ge 0$ and $N_2(t), t ge 0$ are independent Poisson processes with rates $λ_1 = 10$ and $λ_2 = 1$.
The amounts of successive type 1 claims are independent exponential
random variables with mean 1000 dollars whereas the amounts from type 2 claims are independent exponential random variables with mean 5000 dollars. A claim for 4000 dollars has just been received; what is the probability it is a type 1 claim?
Here's my approach to the problem:
beginalign
& P(textclaim = texttype 1mid 4000) \[10pt]
= & fracP(4000mid textclaim = texttype 1)(textclaim = texttype 1)P(4000mid textclaim = texttype 1)P(textclaim = texttype 1) + P(4000midtextclaim = texttype 2)P(textclaim = texttype 2)
endalign
by Bayes' formula.
My question is, how do I calculate $P(4000mid textclaim = texttype 1)$ and $P(4000mid textclaim = texttype 2)$? The dollar value of claims is exponentially distributed, and since the distribution is continuous I can't fix a value to the pdf, I need a range. Any tips on how to calculate these two values?
probability statistics bayes-theorem exponential-distribution
edited Jul 25 at 3:12
Michael Hardy
204k23186461
asked Jul 25 at 3:02
Andrew Louis
656
There are two types of claims that are made to an insurance company. Let $N_i(t)$ denote the number of type $i$ claims made by time $t$, and suppose that $N_1(t), t ge 0$ and $N_2(t), t ge 0$ are independent Poisson processes with rates $λ_1 = 10$ and $λ_2 = 1$.
The amounts of successive type 1 claims are independent exponential
random variables with mean 1000 dollars whereas the amounts from type 2 claims are independent exponential random variables with mean 5000 dollars. A claim for 4000 dollars has just been received; what is the probability it is a type 1 claim?
Here's my approach to the problem:
beginalign
& P(textclaim = texttype 1mid 4000) \[10pt]
= & fracP(4000mid textclaim = texttype 1)(textclaim = texttype 1)P(4000mid textclaim = texttype 1)P(textclaim = texttype 1) + P(4000midtextclaim = texttype 2)P(textclaim = texttype 2)
endalign
by Bayes' formula.
My question is, how do I calculate $P(4000mid textclaim = texttype 1)$ and $P(4000mid textclaim = texttype 2)$? The dollar value of claims is exponentially distributed, and since the distribution is continuous I can't fix a value to the pdf, I need a range. Any tips on how to calculate these two values?
probability statistics bayes-theorem exponential-distribution
edited Jul 25 at 3:12
Michael Hardy
204k23186461
asked Jul 25 at 3:02
Andrew Louis
656
edited Jul 25 at 3:12
Michael Hardy
204k23186461
edited Jul 25 at 3:12
Michael Hardy
204k23186461
edited Jul 25 at 3:12
Michael Hardy
204k23186461
204k23186461
asked Jul 25 at 3:02
Andrew Louis
656
asked Jul 25 at 3:02
Andrew Louis
656
asked Jul 25 at 3:02
Andrew Louis
656
656
It is pretty common to encounter continuous distributions in Bayes distributions. Just take the PDF value. It will do because the range cancels out from the numerator and denominator.
– kasa
Jul 25 at 3:12
add a comment |Â
It is pretty common to encounter continuous distributions in Bayes distributions. Just take the PDF value. It will do because the range cancels out from the numerator and denominator.
– kasa
Jul 25 at 3:12
It is pretty common to encounter continuous distributions in Bayes distributions. Just take the PDF value. It will do because the range cancels out from the numerator and denominator.
– kasa
Jul 25 at 3:12
It is pretty common to encounter continuous distributions in Bayes distributions. Just take the PDF value. It will do because the range cancels out from the numerator and denominator.
– kasa
Jul 25 at 3:12
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Since the claim size has a continuous distribution, the probability that it is a particular amount is $0.$ One uses a probability density instead of a probability in such cases.
beginalign
Pr(texttype 1) & = frac1011 \[10pt]
Pr(texttype 2) & = frac 1 11 \[10pt]
L(texttype 1) & = frac 1 1000 e^-4000/1000 \[10pt]
L(texttype 2) & = frac 1 5000 e^-4000/5000
endalign
$L$ is the likelihood function.
$$
Pr(texttype 1mid 4000) = fracPr(texttype 1)cdot L(texttype 1)Pr(texttype 1)cdot L(texttype 1) + Pr(texttype 2)cdot L(texttype 2).
$$
answered Jul 25 at 3:56
Michael Hardy
204k23186461
add a comment |Â
up vote
1
down vote
Let $T_i,n$ be the arrival times of $N_i(t)$ for and $C_i,n$ the claim values associated with these arrivals, for $i=1,2$. Let $lambda_i$ be the rate of $N_i(t)$ and $1/mu_i$ the mean of $C_i,1$, for $i=1,2$. Let $T_n$ be the superposition of $T_1,n$ and $T_2,n$ and $C_n$ the superposition of $C_1,n$ and $C_2,n$. Then by Bayes' rule we have for any $c>0$
$$
mathbb P(T_1=T_1,1mid C_1 = c) = fracmathbb P(C_1=cmid T_1=T_1,1)mathbb P(T_1=T_1,1)mathbb P(C_1=c).
$$
Now,
$$
mathbb P(T_1=T_i,1) = mathbb P(T_i,1<T_j,1) = fraclambda_ilambda_i+lambda_j, quad (i,j)in(1,2),(2,1),
$$
$$
mathbb P(C_i=cmid T_1 = T_i,1) = f_C_i,1(c) = mu_i e^-mu_i c,quad iin1,2,
$$
and
$$
mathbb P(C_1=c) = mathbb P(C_1=cmid T_1=T_1,1) + mathbb P(C_1=cmid T_1=T_2,1),
$$
and hence
beginalign
mathbb P(T_1=T_i,1) &= fracmu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right)mu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right) + mu_2 e^-mu_2 cleft(fraclambda_2lambda_1+lambda_2right)\
&= fraclambda_1mu_1 e^-mu_1 clambda_1mu_1 e^-mu_1 c + lambda_2mu_2 e^-mu_2 c.
endalign
In this example, we have
beginalign
lambda_1 &= 10\
lambda_2 &= 1\
mu_1 &= 1/1000\
mu_2 &= 1/5000\
c &= 4000,
endalign
and substituting these values yields
$$
frac5050 + e^16/5 approx 0.670848.
$$
answered Jul 25 at 4:11
Math1000
18.4k31444
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Since the claim size has a continuous distribution, the probability that it is a particular amount is $0.$ One uses a probability density instead of a probability in such cases.
beginalign
Pr(texttype 1) & = frac1011 \[10pt]
Pr(texttype 2) & = frac 1 11 \[10pt]
L(texttype 1) & = frac 1 1000 e^-4000/1000 \[10pt]
L(texttype 2) & = frac 1 5000 e^-4000/5000
endalign
$L$ is the likelihood function.
$$
Pr(texttype 1mid 4000) = fracPr(texttype 1)cdot L(texttype 1)Pr(texttype 1)cdot L(texttype 1) + Pr(texttype 2)cdot L(texttype 2).
$$
answered Jul 25 at 3:56
Michael Hardy
204k23186461
add a comment |Â
up vote
1
down vote
Since the claim size has a continuous distribution, the probability that it is a particular amount is $0.$ One uses a probability density instead of a probability in such cases.
beginalign
Pr(texttype 1) & = frac1011 \[10pt]
Pr(texttype 2) & = frac 1 11 \[10pt]
L(texttype 1) & = frac 1 1000 e^-4000/1000 \[10pt]
L(texttype 2) & = frac 1 5000 e^-4000/5000
endalign
$L$ is the likelihood function.
$$
Pr(texttype 1mid 4000) = fracPr(texttype 1)cdot L(texttype 1)Pr(texttype 1)cdot L(texttype 1) + Pr(texttype 2)cdot L(texttype 2).
$$
answered Jul 25 at 3:56
Michael Hardy
204k23186461
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since the claim size has a continuous distribution, the probability that it is a particular amount is $0.$ One uses a probability density instead of a probability in such cases.
beginalign
Pr(texttype 1) & = frac1011 \[10pt]
Pr(texttype 2) & = frac 1 11 \[10pt]
L(texttype 1) & = frac 1 1000 e^-4000/1000 \[10pt]
L(texttype 2) & = frac 1 5000 e^-4000/5000
endalign
$L$ is the likelihood function.
$$
Pr(texttype 1mid 4000) = fracPr(texttype 1)cdot L(texttype 1)Pr(texttype 1)cdot L(texttype 1) + Pr(texttype 2)cdot L(texttype 2).
$$
answered Jul 25 at 3:56
Michael Hardy
204k23186461
Since the claim size has a continuous distribution, the probability that it is a particular amount is $0.$ One uses a probability density instead of a probability in such cases.
beginalign
Pr(texttype 1) & = frac1011 \[10pt]
Pr(texttype 2) & = frac 1 11 \[10pt]
L(texttype 1) & = frac 1 1000 e^-4000/1000 \[10pt]
L(texttype 2) & = frac 1 5000 e^-4000/5000
endalign
$L$ is the likelihood function.
$$
Pr(texttype 1mid 4000) = fracPr(texttype 1)cdot L(texttype 1)Pr(texttype 1)cdot L(texttype 1) + Pr(texttype 2)cdot L(texttype 2).
$$
answered Jul 25 at 3:56
Michael Hardy
204k23186461
answered Jul 25 at 3:56
Michael Hardy
204k23186461
answered Jul 25 at 3:56
Michael Hardy
204k23186461
answered Jul 25 at 3:56
Michael Hardy
204k23186461
204k23186461
add a comment |Â
add a comment |Â
up vote
1
down vote
Let $T_i,n$ be the arrival times of $N_i(t)$ for and $C_i,n$ the claim values associated with these arrivals, for $i=1,2$. Let $lambda_i$ be the rate of $N_i(t)$ and $1/mu_i$ the mean of $C_i,1$, for $i=1,2$. Let $T_n$ be the superposition of $T_1,n$ and $T_2,n$ and $C_n$ the superposition of $C_1,n$ and $C_2,n$. Then by Bayes' rule we have for any $c>0$
$$
mathbb P(T_1=T_1,1mid C_1 = c) = fracmathbb P(C_1=cmid T_1=T_1,1)mathbb P(T_1=T_1,1)mathbb P(C_1=c).
$$
Now,
$$
mathbb P(T_1=T_i,1) = mathbb P(T_i,1<T_j,1) = fraclambda_ilambda_i+lambda_j, quad (i,j)in(1,2),(2,1),
$$
$$
mathbb P(C_i=cmid T_1 = T_i,1) = f_C_i,1(c) = mu_i e^-mu_i c,quad iin1,2,
$$
and
$$
mathbb P(C_1=c) = mathbb P(C_1=cmid T_1=T_1,1) + mathbb P(C_1=cmid T_1=T_2,1),
$$
and hence
beginalign
mathbb P(T_1=T_i,1) &= fracmu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right)mu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right) + mu_2 e^-mu_2 cleft(fraclambda_2lambda_1+lambda_2right)\
&= fraclambda_1mu_1 e^-mu_1 clambda_1mu_1 e^-mu_1 c + lambda_2mu_2 e^-mu_2 c.
endalign
In this example, we have
beginalign
lambda_1 &= 10\
lambda_2 &= 1\
mu_1 &= 1/1000\
mu_2 &= 1/5000\
c &= 4000,
endalign
and substituting these values yields
$$
frac5050 + e^16/5 approx 0.670848.
$$
answered Jul 25 at 4:11
Math1000
18.4k31444
add a comment |Â
up vote
1
down vote
Let $T_i,n$ be the arrival times of $N_i(t)$ for and $C_i,n$ the claim values associated with these arrivals, for $i=1,2$. Let $lambda_i$ be the rate of $N_i(t)$ and $1/mu_i$ the mean of $C_i,1$, for $i=1,2$. Let $T_n$ be the superposition of $T_1,n$ and $T_2,n$ and $C_n$ the superposition of $C_1,n$ and $C_2,n$. Then by Bayes' rule we have for any $c>0$
$$
mathbb P(T_1=T_1,1mid C_1 = c) = fracmathbb P(C_1=cmid T_1=T_1,1)mathbb P(T_1=T_1,1)mathbb P(C_1=c).
$$
Now,
$$
mathbb P(T_1=T_i,1) = mathbb P(T_i,1<T_j,1) = fraclambda_ilambda_i+lambda_j, quad (i,j)in(1,2),(2,1),
$$
$$
mathbb P(C_i=cmid T_1 = T_i,1) = f_C_i,1(c) = mu_i e^-mu_i c,quad iin1,2,
$$
and
$$
mathbb P(C_1=c) = mathbb P(C_1=cmid T_1=T_1,1) + mathbb P(C_1=cmid T_1=T_2,1),
$$
and hence
beginalign
mathbb P(T_1=T_i,1) &= fracmu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right)mu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right) + mu_2 e^-mu_2 cleft(fraclambda_2lambda_1+lambda_2right)\
&= fraclambda_1mu_1 e^-mu_1 clambda_1mu_1 e^-mu_1 c + lambda_2mu_2 e^-mu_2 c.
endalign
In this example, we have
beginalign
lambda_1 &= 10\
lambda_2 &= 1\
mu_1 &= 1/1000\
mu_2 &= 1/5000\
c &= 4000,
endalign
and substituting these values yields
$$
frac5050 + e^16/5 approx 0.670848.
$$
answered Jul 25 at 4:11
Math1000
18.4k31444
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $T_i,n$ be the arrival times of $N_i(t)$ for and $C_i,n$ the claim values associated with these arrivals, for $i=1,2$. Let $lambda_i$ be the rate of $N_i(t)$ and $1/mu_i$ the mean of $C_i,1$, for $i=1,2$. Let $T_n$ be the superposition of $T_1,n$ and $T_2,n$ and $C_n$ the superposition of $C_1,n$ and $C_2,n$. Then by Bayes' rule we have for any $c>0$
$$
mathbb P(T_1=T_1,1mid C_1 = c) = fracmathbb P(C_1=cmid T_1=T_1,1)mathbb P(T_1=T_1,1)mathbb P(C_1=c).
$$
Now,
$$
mathbb P(T_1=T_i,1) = mathbb P(T_i,1<T_j,1) = fraclambda_ilambda_i+lambda_j, quad (i,j)in(1,2),(2,1),
$$
$$
mathbb P(C_i=cmid T_1 = T_i,1) = f_C_i,1(c) = mu_i e^-mu_i c,quad iin1,2,
$$
and
$$
mathbb P(C_1=c) = mathbb P(C_1=cmid T_1=T_1,1) + mathbb P(C_1=cmid T_1=T_2,1),
$$
and hence
beginalign
mathbb P(T_1=T_i,1) &= fracmu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right)mu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right) + mu_2 e^-mu_2 cleft(fraclambda_2lambda_1+lambda_2right)\
&= fraclambda_1mu_1 e^-mu_1 clambda_1mu_1 e^-mu_1 c + lambda_2mu_2 e^-mu_2 c.
endalign
In this example, we have
beginalign
lambda_1 &= 10\
lambda_2 &= 1\
mu_1 &= 1/1000\
mu_2 &= 1/5000\
c &= 4000,
endalign
and substituting these values yields
$$
frac5050 + e^16/5 approx 0.670848.
$$
answered Jul 25 at 4:11
Math1000
18.4k31444
Let $T_i,n$ be the arrival times of $N_i(t)$ for and $C_i,n$ the claim values associated with these arrivals, for $i=1,2$. Let $lambda_i$ be the rate of $N_i(t)$ and $1/mu_i$ the mean of $C_i,1$, for $i=1,2$. Let $T_n$ be the superposition of $T_1,n$ and $T_2,n$ and $C_n$ the superposition of $C_1,n$ and $C_2,n$. Then by Bayes' rule we have for any $c>0$
$$
mathbb P(T_1=T_1,1mid C_1 = c) = fracmathbb P(C_1=cmid T_1=T_1,1)mathbb P(T_1=T_1,1)mathbb P(C_1=c).
$$
Now,
$$
mathbb P(T_1=T_i,1) = mathbb P(T_i,1<T_j,1) = fraclambda_ilambda_i+lambda_j, quad (i,j)in(1,2),(2,1),
$$
$$
mathbb P(C_i=cmid T_1 = T_i,1) = f_C_i,1(c) = mu_i e^-mu_i c,quad iin1,2,
$$
and
$$
mathbb P(C_1=c) = mathbb P(C_1=cmid T_1=T_1,1) + mathbb P(C_1=cmid T_1=T_2,1),
$$
and hence
beginalign
mathbb P(T_1=T_i,1) &= fracmu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right)mu_1 e^-mu_1 cleft(fraclambda_1lambda_1+lambda_2right) + mu_2 e^-mu_2 cleft(fraclambda_2lambda_1+lambda_2right)\
&= fraclambda_1mu_1 e^-mu_1 clambda_1mu_1 e^-mu_1 c + lambda_2mu_2 e^-mu_2 c.
endalign
In this example, we have
beginalign
lambda_1 &= 10\
lambda_2 &= 1\
mu_1 &= 1/1000\
mu_2 &= 1/5000\
c &= 4000,
endalign
and substituting these values yields
$$
frac5050 + e^16/5 approx 0.670848.
$$
answered Jul 25 at 4:11
Math1000
18.4k31444
answered Jul 25 at 4:11
Math1000
18.4k31444
answered Jul 25 at 4:11
Math1000
18.4k31444
answered Jul 25 at 4:11
Math1000
18.4k31444
18.4k31444
add a comment |Â
add a comment |Â
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It is pretty common to encounter continuous distributions in Bayes distributions. Just take the PDF value. It will do because the range cancels out from the numerator and denominator.
– kasa
Jul 25 at 3:12