System of two electronic devices in parallel, with conditional lifetimes.
$begingroup$
Suppose we have a system made up of two elements in parallel. I know that their lifetime, conditioned to $Theta = lambda$, are independent exponential with parameters respectively $lambda$ and $2lambda$. The density of $Theta$ is an exponential of parameter 1.
I want to calculate the probability that the system lasts for at least a time t.
Below is what I have done, I feel I am very close to the solution but I cannot work out where to integrate.
Let $T_1$ and $T_2$ be the lifetimes of the two devices.
By properties of conditional probability, I can write this:
$$max(T_1,T_2) = [max(T_1,T_2)vert lambda] *f(lambda)$$
Where $max(T_1,T_2)$ would be the joint density function.
Now I feel like I can write this:
$$mathbb{P}[max(T_1,T_2)>t] = int_?^?mathbb{P}[max(T_1,T_2)>tvert lambda] f(lambda)dlambda$$
However I am not sure what to put instead of the two ?.
Could you shed some light on that?
probability conditional-probability
asked Jan 7 at 8:49
qcc101qcc101
629213
$endgroup$
add a comment |
$begingroup$
Suppose we have a system made up of two elements in parallel. I know that their lifetime, conditioned to $Theta = lambda$, are independent exponential with parameters respectively $lambda$ and $2lambda$. The density of $Theta$ is an exponential of parameter 1.
I want to calculate the probability that the system lasts for at least a time t.
Below is what I have done, I feel I am very close to the solution but I cannot work out where to integrate.
Let $T_1$ and $T_2$ be the lifetimes of the two devices.
By properties of conditional probability, I can write this:
$$max(T_1,T_2) = [max(T_1,T_2)vert lambda] *f(lambda)$$
Where $max(T_1,T_2)$ would be the joint density function.
Now I feel like I can write this:
$$mathbb{P}[max(T_1,T_2)>t] = int_?^?mathbb{P}[max(T_1,T_2)>tvert lambda] f(lambda)dlambda$$
However I am not sure what to put instead of the two ?.
Could you shed some light on that?
probability conditional-probability
asked Jan 7 at 8:49
qcc101qcc101
629213
$endgroup$
$begingroup$
In general the integration limits is $-infty$ and $infty$. In your case since $Theta$ is exponential, the support is $(0, infty)$, so you can replace with that. (i.e. $f(lambda) = 0$ when $lambda leq 0$.)
$endgroup$
– BGM
Jan 7 at 10:10
add a comment |
$begingroup$
Suppose we have a system made up of two elements in parallel. I know that their lifetime, conditioned to $Theta = lambda$, are independent exponential with parameters respectively $lambda$ and $2lambda$. The density of $Theta$ is an exponential of parameter 1.
I want to calculate the probability that the system lasts for at least a time t.
Below is what I have done, I feel I am very close to the solution but I cannot work out where to integrate.
Let $T_1$ and $T_2$ be the lifetimes of the two devices.
By properties of conditional probability, I can write this:
$$max(T_1,T_2) = [max(T_1,T_2)vert lambda] *f(lambda)$$
Where $max(T_1,T_2)$ would be the joint density function.
Now I feel like I can write this:
$$mathbb{P}[max(T_1,T_2)>t] = int_?^?mathbb{P}[max(T_1,T_2)>tvert lambda] f(lambda)dlambda$$
However I am not sure what to put instead of the two ?.
Could you shed some light on that?
probability conditional-probability
asked Jan 7 at 8:49
qcc101qcc101
629213
$endgroup$
Suppose we have a system made up of two elements in parallel. I know that their lifetime, conditioned to $Theta = lambda$, are independent exponential with parameters respectively $lambda$ and $2lambda$. The density of $Theta$ is an exponential of parameter 1.
I want to calculate the probability that the system lasts for at least a time t.
Below is what I have done, I feel I am very close to the solution but I cannot work out where to integrate.
Let $T_1$ and $T_2$ be the lifetimes of the two devices.
By properties of conditional probability, I can write this:
$$max(T_1,T_2) = [max(T_1,T_2)vert lambda] *f(lambda)$$
Where $max(T_1,T_2)$ would be the joint density function.
Now I feel like I can write this:
$$mathbb{P}[max(T_1,T_2)>t] = int_?^?mathbb{P}[max(T_1,T_2)>tvert lambda] f(lambda)dlambda$$
However I am not sure what to put instead of the two ?.
Could you shed some light on that?
probability conditional-probability
probability conditional-probability
asked Jan 7 at 8:49
qcc101qcc101
629213
asked Jan 7 at 8:49
qcc101qcc101
629213
edited Jan 7 at 9:15
qcc101
asked Jan 7 at 8:49
qcc101qcc101
629213
asked Jan 7 at 8:49
qcc101qcc101
629213
asked Jan 7 at 8:49
qcc101qcc101
629213
629213
$begingroup$
In general the integration limits is $-infty$ and $infty$. In your case since $Theta$ is exponential, the support is $(0, infty)$, so you can replace with that. (i.e. $f(lambda) = 0$ when $lambda leq 0$.)
$endgroup$
– BGM
Jan 7 at 10:10
add a comment |
$begingroup$
In general the integration limits is $-infty$ and $infty$. In your case since $Theta$ is exponential, the support is $(0, infty)$, so you can replace with that. (i.e. $f(lambda) = 0$ when $lambda leq 0$.)
$endgroup$
– BGM
Jan 7 at 10:10
$begingroup$
In general the integration limits is $-infty$ and $infty$. In your case since $Theta$ is exponential, the support is $(0, infty)$, so you can replace with that. (i.e. $f(lambda) = 0$ when $lambda leq 0$.)
$endgroup$
– BGM
Jan 7 at 10:10
$begingroup$
In general the integration limits is $-infty$ and $infty$. In your case since $Theta$ is exponential, the support is $(0, infty)$, so you can replace with that. (i.e. $f(lambda) = 0$ when $lambda leq 0$.)
$endgroup$
– BGM
Jan 7 at 10:10
add a comment |
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$begingroup$
In general the integration limits is $-infty$ and $infty$. In your case since $Theta$ is exponential, the support is $(0, infty)$, so you can replace with that. (i.e. $f(lambda) = 0$ when $lambda leq 0$.)
$endgroup$
– BGM
Jan 7 at 10:10