independent Poisson processes probability
$begingroup$
I have been given this question to solve
The numbers of claims to an insurance company from smokers and nonsmokers follow independent Poisson processes. On average 4 claims from nonsmokers and 6 claims from smokers arrive every day independently of each
other.
Given that 8 claims arrived in a day, what is the probability that 5 of them
were from smokers?
so my attempt is:
$X_{t} :$ number of claims from non-smokers (with intensity $lambda = 4$)
$Y_{t} :$ number of claims from smokers (with intensity $alpha = 6$)
So from this I have
$Z_{t}:$ number of claims from smokers & non-smokers (with intensity $lambda+alpha = 10$)
So I have set the probability to be
$$frac{P(X_{1}=3)P(Y_{1}=5)}{P(Z_{1}= 8)}$$
I'm not sure if this is correct, I'm uncertain if I need to include $X_{t}$ in the equation.
probability stochastic-processes poisson-process
asked Dec 18 '18 at 14:03
Rito LoweRito Lowe
465
$endgroup$
add a comment |
$begingroup$
I have been given this question to solve
The numbers of claims to an insurance company from smokers and nonsmokers follow independent Poisson processes. On average 4 claims from nonsmokers and 6 claims from smokers arrive every day independently of each
other.
Given that 8 claims arrived in a day, what is the probability that 5 of them
were from smokers?
so my attempt is:
$X_{t} :$ number of claims from non-smokers (with intensity $lambda = 4$)
$Y_{t} :$ number of claims from smokers (with intensity $alpha = 6$)
So from this I have
$Z_{t}:$ number of claims from smokers & non-smokers (with intensity $lambda+alpha = 10$)
So I have set the probability to be
$$frac{P(X_{1}=3)P(Y_{1}=5)}{P(Z_{1}= 8)}$$
I'm not sure if this is correct, I'm uncertain if I need to include $X_{t}$ in the equation.
probability stochastic-processes poisson-process
asked Dec 18 '18 at 14:03
Rito LoweRito Lowe
465
$endgroup$
$begingroup$
This seems correct. You calculate $P(Y_1 = 5 | Z_1 = 8) = frac{P({Y_1 = 5}cap{X_1 + Y_1 = 8})}{P(Z_1 = 8)} = frac{P({Y_1 = 5}cap{X_1 = 3})}{P(Z_1 = 8)} = frac{P(Y_1 = 5)P(X_1= 3)}{P(Z_1 = 8)}$, so yes, you need to include $X_1$.
$endgroup$
– Tki Deneb
Dec 18 '18 at 15:06
add a comment |
$begingroup$
I have been given this question to solve
The numbers of claims to an insurance company from smokers and nonsmokers follow independent Poisson processes. On average 4 claims from nonsmokers and 6 claims from smokers arrive every day independently of each
other.
Given that 8 claims arrived in a day, what is the probability that 5 of them
were from smokers?
so my attempt is:
$X_{t} :$ number of claims from non-smokers (with intensity $lambda = 4$)
$Y_{t} :$ number of claims from smokers (with intensity $alpha = 6$)
So from this I have
$Z_{t}:$ number of claims from smokers & non-smokers (with intensity $lambda+alpha = 10$)
So I have set the probability to be
$$frac{P(X_{1}=3)P(Y_{1}=5)}{P(Z_{1}= 8)}$$
I'm not sure if this is correct, I'm uncertain if I need to include $X_{t}$ in the equation.
probability stochastic-processes poisson-process
asked Dec 18 '18 at 14:03
Rito LoweRito Lowe
465
$endgroup$
I have been given this question to solve
The numbers of claims to an insurance company from smokers and nonsmokers follow independent Poisson processes. On average 4 claims from nonsmokers and 6 claims from smokers arrive every day independently of each
other.
Given that 8 claims arrived in a day, what is the probability that 5 of them
were from smokers?
so my attempt is:
$X_{t} :$ number of claims from non-smokers (with intensity $lambda = 4$)
$Y_{t} :$ number of claims from smokers (with intensity $alpha = 6$)
So from this I have
$Z_{t}:$ number of claims from smokers & non-smokers (with intensity $lambda+alpha = 10$)
So I have set the probability to be
$$frac{P(X_{1}=3)P(Y_{1}=5)}{P(Z_{1}= 8)}$$
I'm not sure if this is correct, I'm uncertain if I need to include $X_{t}$ in the equation.
probability stochastic-processes poisson-process
probability stochastic-processes poisson-process
asked Dec 18 '18 at 14:03
Rito LoweRito Lowe
465
asked Dec 18 '18 at 14:03
Rito LoweRito Lowe
465
edited Dec 18 '18 at 14:07
Rito Lowe
asked Dec 18 '18 at 14:03
Rito LoweRito Lowe
465
asked Dec 18 '18 at 14:03
Rito LoweRito Lowe
465
asked Dec 18 '18 at 14:03
Rito LoweRito Lowe
465
465
$begingroup$
This seems correct. You calculate $P(Y_1 = 5 | Z_1 = 8) = frac{P({Y_1 = 5}cap{X_1 + Y_1 = 8})}{P(Z_1 = 8)} = frac{P({Y_1 = 5}cap{X_1 = 3})}{P(Z_1 = 8)} = frac{P(Y_1 = 5)P(X_1= 3)}{P(Z_1 = 8)}$, so yes, you need to include $X_1$.
$endgroup$
– Tki Deneb
Dec 18 '18 at 15:06
add a comment |
$begingroup$
This seems correct. You calculate $P(Y_1 = 5 | Z_1 = 8) = frac{P({Y_1 = 5}cap{X_1 + Y_1 = 8})}{P(Z_1 = 8)} = frac{P({Y_1 = 5}cap{X_1 = 3})}{P(Z_1 = 8)} = frac{P(Y_1 = 5)P(X_1= 3)}{P(Z_1 = 8)}$, so yes, you need to include $X_1$.
$endgroup$
– Tki Deneb
Dec 18 '18 at 15:06
$begingroup$
This seems correct. You calculate $P(Y_1 = 5 | Z_1 = 8) = frac{P({Y_1 = 5}cap{X_1 + Y_1 = 8})}{P(Z_1 = 8)} = frac{P({Y_1 = 5}cap{X_1 = 3})}{P(Z_1 = 8)} = frac{P(Y_1 = 5)P(X_1= 3)}{P(Z_1 = 8)}$, so yes, you need to include $X_1$.
$endgroup$
– Tki Deneb
Dec 18 '18 at 15:06
$begingroup$
This seems correct. You calculate $P(Y_1 = 5 | Z_1 = 8) = frac{P({Y_1 = 5}cap{X_1 + Y_1 = 8})}{P(Z_1 = 8)} = frac{P({Y_1 = 5}cap{X_1 = 3})}{P(Z_1 = 8)} = frac{P(Y_1 = 5)P(X_1= 3)}{P(Z_1 = 8)}$, so yes, you need to include $X_1$.
$endgroup$
– Tki Deneb
Dec 18 '18 at 15:06
add a comment |
1 Answer
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$begingroup$
$P(Y_1=5|Z_1=8)$
$ = frac{P(Y_1=5,; Z_1=8)}{P(Z_1=8)}$ (by Baye's rule)
$ = frac{P(Y_1=5, ; X_1=3)}{P(Z_1=8)}$ (since $Z_1=X_1+Y_1$)
$ = frac{P(Y_1=5)P(X_1=3)}{P(Z_1=8)}$ (since $X_1$ and $Y_1$ are independent)
Since $X_1,Y_1$, and $Z_1$ are all Poisson, you can easily plug in the standard formula in the above equation
answered Dec 18 '18 at 20:22
Aditya DuaAditya Dua
1,00418
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$P(Y_1=5|Z_1=8)$
$ = frac{P(Y_1=5,; Z_1=8)}{P(Z_1=8)}$ (by Baye's rule)
$ = frac{P(Y_1=5, ; X_1=3)}{P(Z_1=8)}$ (since $Z_1=X_1+Y_1$)
$ = frac{P(Y_1=5)P(X_1=3)}{P(Z_1=8)}$ (since $X_1$ and $Y_1$ are independent)
Since $X_1,Y_1$, and $Z_1$ are all Poisson, you can easily plug in the standard formula in the above equation
answered Dec 18 '18 at 20:22
Aditya DuaAditya Dua
1,00418
$endgroup$
add a comment |
$begingroup$
$P(Y_1=5|Z_1=8)$
$ = frac{P(Y_1=5,; Z_1=8)}{P(Z_1=8)}$ (by Baye's rule)
$ = frac{P(Y_1=5, ; X_1=3)}{P(Z_1=8)}$ (since $Z_1=X_1+Y_1$)
$ = frac{P(Y_1=5)P(X_1=3)}{P(Z_1=8)}$ (since $X_1$ and $Y_1$ are independent)
Since $X_1,Y_1$, and $Z_1$ are all Poisson, you can easily plug in the standard formula in the above equation
answered Dec 18 '18 at 20:22
Aditya DuaAditya Dua
1,00418
$endgroup$
add a comment |
$begingroup$
$P(Y_1=5|Z_1=8)$
$ = frac{P(Y_1=5,; Z_1=8)}{P(Z_1=8)}$ (by Baye's rule)
$ = frac{P(Y_1=5, ; X_1=3)}{P(Z_1=8)}$ (since $Z_1=X_1+Y_1$)
$ = frac{P(Y_1=5)P(X_1=3)}{P(Z_1=8)}$ (since $X_1$ and $Y_1$ are independent)
Since $X_1,Y_1$, and $Z_1$ are all Poisson, you can easily plug in the standard formula in the above equation
answered Dec 18 '18 at 20:22
Aditya DuaAditya Dua
1,00418
$endgroup$
$P(Y_1=5|Z_1=8)$
$ = frac{P(Y_1=5,; Z_1=8)}{P(Z_1=8)}$ (by Baye's rule)
$ = frac{P(Y_1=5, ; X_1=3)}{P(Z_1=8)}$ (since $Z_1=X_1+Y_1$)
$ = frac{P(Y_1=5)P(X_1=3)}{P(Z_1=8)}$ (since $X_1$ and $Y_1$ are independent)
Since $X_1,Y_1$, and $Z_1$ are all Poisson, you can easily plug in the standard formula in the above equation
answered Dec 18 '18 at 20:22
Aditya DuaAditya Dua
1,00418
answered Dec 18 '18 at 20:22
Aditya DuaAditya Dua
1,00418
answered Dec 18 '18 at 20:22
Aditya DuaAditya Dua
1,00418
answered Dec 18 '18 at 20:22
Aditya DuaAditya Dua
1,00418
1,00418
add a comment |
add a comment |
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$begingroup$
This seems correct. You calculate $P(Y_1 = 5 | Z_1 = 8) = frac{P({Y_1 = 5}cap{X_1 + Y_1 = 8})}{P(Z_1 = 8)} = frac{P({Y_1 = 5}cap{X_1 = 3})}{P(Z_1 = 8)} = frac{P(Y_1 = 5)P(X_1= 3)}{P(Z_1 = 8)}$, so yes, you need to include $X_1$.
$endgroup$
– Tki Deneb
Dec 18 '18 at 15:06