Time dependent probabilities
I have a problem understanding how probabilities develop over time. An example:
A radioactive element has a half life $t_{1/2}$.
That must mean that for an individual atom the probability that it has decayed at $t_{1/2}$ is $0.5$. (Sorry about the ugly notation $t_{1/2}$ means half life, I change to $k$ henceforth to make it look a bit cleaner)
After two half lives only a quarter of the original population remains.
I interpret this as the cumulative probability $P(t)$ of an atom having decayed after a time $t$ should look like $P(t)=1-0.5^{frac{t}{k}}$. This makes sense with $P(0)=0$ and $P(inf)=1$.
Now for the part where I get confused: There must be such a thing as an instantaneous probability at every moment that integrates to $P(t)$. You'd guess that this should be the derivative $frac{dP}{dt}=frac{ln(2) cdot 0.5^{t/k}}{k}$. However, this probability is time dependent and that is unphysical. The atom must have a constant probability of decay at any given moment. It doesn't know how I set up the experiment. I understand that there is something about probabilities that I don't understand. Any insight would be appreciated.
probability
edited 2 days ago
amWhy
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asked 2 days ago
OskarOskar
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I have a problem understanding how probabilities develop over time. An example:
A radioactive element has a half life $t_{1/2}$.
That must mean that for an individual atom the probability that it has decayed at $t_{1/2}$ is $0.5$. (Sorry about the ugly notation $t_{1/2}$ means half life, I change to $k$ henceforth to make it look a bit cleaner)
After two half lives only a quarter of the original population remains.
I interpret this as the cumulative probability $P(t)$ of an atom having decayed after a time $t$ should look like $P(t)=1-0.5^{frac{t}{k}}$. This makes sense with $P(0)=0$ and $P(inf)=1$.
Now for the part where I get confused: There must be such a thing as an instantaneous probability at every moment that integrates to $P(t)$. You'd guess that this should be the derivative $frac{dP}{dt}=frac{ln(2) cdot 0.5^{t/k}}{k}$. However, this probability is time dependent and that is unphysical. The atom must have a constant probability of decay at any given moment. It doesn't know how I set up the experiment. I understand that there is something about probabilities that I don't understand. Any insight would be appreciated.
probability
edited 2 days ago
amWhy
192k28225439
asked 2 days ago
OskarOskar
1
New contributor
Oskar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
– Oskar
2 days ago
You missed that $P'(t)=ae^{-at}$ is the variation of the total proportion of decayed atoms at time $t$. The probability of a single non decayed atom at time $t$ to decay between times $t$ and $t+dt$ is $acdot dt$.
– Did
2 days ago
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
– Oskar
2 days ago
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
– Did
2 days ago
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
– Oskar
yesterday
|
show 3 more comments
I have a problem understanding how probabilities develop over time. An example:
A radioactive element has a half life $t_{1/2}$.
That must mean that for an individual atom the probability that it has decayed at $t_{1/2}$ is $0.5$. (Sorry about the ugly notation $t_{1/2}$ means half life, I change to $k$ henceforth to make it look a bit cleaner)
After two half lives only a quarter of the original population remains.
I interpret this as the cumulative probability $P(t)$ of an atom having decayed after a time $t$ should look like $P(t)=1-0.5^{frac{t}{k}}$. This makes sense with $P(0)=0$ and $P(inf)=1$.
Now for the part where I get confused: There must be such a thing as an instantaneous probability at every moment that integrates to $P(t)$. You'd guess that this should be the derivative $frac{dP}{dt}=frac{ln(2) cdot 0.5^{t/k}}{k}$. However, this probability is time dependent and that is unphysical. The atom must have a constant probability of decay at any given moment. It doesn't know how I set up the experiment. I understand that there is something about probabilities that I don't understand. Any insight would be appreciated.
probability
edited 2 days ago
amWhy
192k28225439
asked 2 days ago
OskarOskar
1
New contributor
Oskar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a problem understanding how probabilities develop over time. An example:
A radioactive element has a half life $t_{1/2}$.
That must mean that for an individual atom the probability that it has decayed at $t_{1/2}$ is $0.5$. (Sorry about the ugly notation $t_{1/2}$ means half life, I change to $k$ henceforth to make it look a bit cleaner)
After two half lives only a quarter of the original population remains.
I interpret this as the cumulative probability $P(t)$ of an atom having decayed after a time $t$ should look like $P(t)=1-0.5^{frac{t}{k}}$. This makes sense with $P(0)=0$ and $P(inf)=1$.
Now for the part where I get confused: There must be such a thing as an instantaneous probability at every moment that integrates to $P(t)$. You'd guess that this should be the derivative $frac{dP}{dt}=frac{ln(2) cdot 0.5^{t/k}}{k}$. However, this probability is time dependent and that is unphysical. The atom must have a constant probability of decay at any given moment. It doesn't know how I set up the experiment. I understand that there is something about probabilities that I don't understand. Any insight would be appreciated.
probability
probability
edited 2 days ago
amWhy
192k28225439
asked 2 days ago
OskarOskar
1
New contributor
Oskar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
amWhy
192k28225439
asked 2 days ago
OskarOskar
1
New contributor
Oskar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
amWhy
192k28225439
edited 2 days ago
amWhy
192k28225439
edited 2 days ago
amWhy
192k28225439
192k28225439
asked 2 days ago
OskarOskar
1
New contributor
Oskar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
OskarOskar
1
asked 2 days ago
OskarOskar
1
1
New contributor
Oskar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Oskar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Oskar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
– Oskar
2 days ago
You missed that $P'(t)=ae^{-at}$ is the variation of the total proportion of decayed atoms at time $t$. The probability of a single non decayed atom at time $t$ to decay between times $t$ and $t+dt$ is $acdot dt$.
– Did
2 days ago
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
– Oskar
2 days ago
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
– Did
2 days ago
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
– Oskar
yesterday
|
show 3 more comments
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
– Oskar
2 days ago
You missed that $P'(t)=ae^{-at}$ is the variation of the total proportion of decayed atoms at time $t$. The probability of a single non decayed atom at time $t$ to decay between times $t$ and $t+dt$ is $acdot dt$.
– Did
2 days ago
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
– Oskar
2 days ago
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
– Did
2 days ago
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
– Oskar
yesterday
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
– Oskar
2 days ago
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
– Oskar
2 days ago
You missed that $P'(t)=ae^{-at}$ is the variation of the total proportion of decayed atoms at time $t$. The probability of a single non decayed atom at time $t$ to decay between times $t$ and $t+dt$ is $acdot dt$.
– Did
2 days ago
You missed that $P'(t)=ae^{-at}$ is the variation of the total proportion of decayed atoms at time $t$. The probability of a single non decayed atom at time $t$ to decay between times $t$ and $t+dt$ is $acdot dt$.
– Did
2 days ago
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
– Oskar
2 days ago
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
– Oskar
2 days ago
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
– Did
2 days ago
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
– Did
2 days ago
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
– Oskar
yesterday
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
– Oskar
yesterday
|
show 3 more comments
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Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
– Oskar
2 days ago
You missed that $P'(t)=ae^{-at}$ is the variation of the total proportion of decayed atoms at time $t$. The probability of a single non decayed atom at time $t$ to decay between times $t$ and $t+dt$ is $acdot dt$.
– Did
2 days ago
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
– Oskar
2 days ago
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
– Did
2 days ago
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
– Oskar
yesterday