Mixing
Time dependent probabilities
$begingroup$
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 Jan 5 at 16:00
amWhy
192k28225439
asked Jan 5 at 15:31
OskarOskar
1
$endgroup$
|
show 5 more comments
$begingroup$
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 Jan 5 at 16:00
amWhy
192k28225439
asked Jan 5 at 15:31
OskarOskar
1
$endgroup$
$begingroup$
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
$endgroup$
– Oskar
Jan 5 at 15:55
$begingroup$
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 $acdot dt$.
$endgroup$
– Did
Jan 5 at 16:28
$begingroup$
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
$endgroup$
– Oskar
Jan 5 at 19:48
$begingroup$
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
$endgroup$
– Did
Jan 6 at 0:25
$begingroup$
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
$endgroup$
– Oskar
Jan 6 at 6:49
|
show 5 more comments
$begingroup$
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 Jan 5 at 16:00
amWhy
192k28225439
asked Jan 5 at 15:31
OskarOskar
1
$endgroup$
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 Jan 5 at 16:00
amWhy
192k28225439
asked Jan 5 at 15:31
OskarOskar
1
edited Jan 5 at 16:00
amWhy
192k28225439
asked Jan 5 at 15:31
OskarOskar
1
edited Jan 5 at 16:00
amWhy
192k28225439
edited Jan 5 at 16:00
amWhy
192k28225439
edited Jan 5 at 16:00
amWhy
192k28225439
192k28225439
asked Jan 5 at 15:31
OskarOskar
1
asked Jan 5 at 15:31
OskarOskar
1
asked Jan 5 at 15:31
OskarOskar
1
1
$begingroup$
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
$endgroup$
– Oskar
Jan 5 at 15:55
$begingroup$
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 $acdot dt$.
$endgroup$
– Did
Jan 5 at 16:28
$begingroup$
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
$endgroup$
– Oskar
Jan 5 at 19:48
$begingroup$
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
$endgroup$
– Did
Jan 6 at 0:25
$begingroup$
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
$endgroup$
– Oskar
Jan 6 at 6:49
|
show 5 more comments
$begingroup$
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
$endgroup$
– Oskar
Jan 5 at 15:55
$begingroup$
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 $acdot dt$.
$endgroup$
– Did
Jan 5 at 16:28
$begingroup$
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
$endgroup$
– Oskar
Jan 5 at 19:48
$begingroup$
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
$endgroup$
– Did
Jan 6 at 0:25
$begingroup$
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
$endgroup$
– Oskar
Jan 6 at 6:49
$begingroup$
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
$endgroup$
– Oskar
Jan 5 at 15:55
$begingroup$
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
$endgroup$
– Oskar
Jan 5 at 15:55
$begingroup$
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 $acdot dt$.
$endgroup$
– Did
Jan 5 at 16:28
$begingroup$
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 $acdot dt$.
$endgroup$
– Did
Jan 5 at 16:28
$begingroup$
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
$endgroup$
– Oskar
Jan 5 at 19:48
$begingroup$
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
$endgroup$
– Oskar
Jan 5 at 19:48
$begingroup$
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
$endgroup$
– Did
Jan 6 at 0:25
$begingroup$
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
$endgroup$
– Did
Jan 6 at 0:25
$begingroup$
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
$endgroup$
– Oskar
Jan 6 at 6:49
$begingroup$
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
$endgroup$
– Oskar
Jan 6 at 6:49
|
show 5 more comments
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$begingroup$
Thanks for the clean-up of the formulas but it should be ln2 x 0.5^(t/k).
$endgroup$
– Oskar
Jan 5 at 15:55
$begingroup$
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 $acdot dt$.
$endgroup$
– Did
Jan 5 at 16:28
$begingroup$
But then the cumulative probability will increase linearly with time and soon be >1. Or am I missing something?
$endgroup$
– Oskar
Jan 5 at 19:48
$begingroup$
No, the cumulative probability increases nonlinearly up to $1$. What makes you think otherwise?
$endgroup$
– Did
Jan 6 at 0:25
$begingroup$
integration of dP=(a x dt) from 0 to t gives P(t)=at, right?
$endgroup$
– Oskar
Jan 6 at 6:49