Definition of Probability mass function of a random process.
A random process $X(t)$ is defined as $X(t)=1, 0leq t leq Y$ and $0$ otherwise where $Y$ follows an exponential distribution. What is the $pmf$ of $X(t)$
I am a bit confused on the definition of $pmf$ for random process. By definition, the pmf of $X(t)$ is
$$Pr({ X(t, xi)=1 })=Pr({ (t, xi) | 0leq t leq xi })=int_0^{infty}int_0^xi f(t,xi)dtdxi$$
However, I am confused how I can calculate $f(t, xi)$. I know from the question that $f(xi) = lambda e^{-lambda xi}$ where $lambda$ is the parameter of the geometric distribution. But I don't see how to get $f(t,xi)$
stochastic-processes
asked Nov 27 '18 at 8:58
AspiringMat
545518
add a comment |
A random process $X(t)$ is defined as $X(t)=1, 0leq t leq Y$ and $0$ otherwise where $Y$ follows an exponential distribution. What is the $pmf$ of $X(t)$
I am a bit confused on the definition of $pmf$ for random process. By definition, the pmf of $X(t)$ is
$$Pr({ X(t, xi)=1 })=Pr({ (t, xi) | 0leq t leq xi })=int_0^{infty}int_0^xi f(t,xi)dtdxi$$
However, I am confused how I can calculate $f(t, xi)$. I know from the question that $f(xi) = lambda e^{-lambda xi}$ where $lambda$ is the parameter of the geometric distribution. But I don't see how to get $f(t,xi)$
stochastic-processes
asked Nov 27 '18 at 8:58
AspiringMat
545518
My guess is that the question asks you to find the distribution of $X(t)$ for a fixed $t$ (and this is trivial). The distribution of the entire process is a measure on an infinite dimensional function space.
– Kavi Rama Murthy
Nov 27 '18 at 9:13
@KaviRamaMurthy That makes a lot more sense now. Just integrate from $t$ to infinity the geometric pdf. This is one of the reason I don't like the notation of suppressing the sample space in random processes. It just makes things more confusing for learners such as myself. Thank you!
– AspiringMat
Nov 27 '18 at 9:22
add a comment |
A random process $X(t)$ is defined as $X(t)=1, 0leq t leq Y$ and $0$ otherwise where $Y$ follows an exponential distribution. What is the $pmf$ of $X(t)$
I am a bit confused on the definition of $pmf$ for random process. By definition, the pmf of $X(t)$ is
$$Pr({ X(t, xi)=1 })=Pr({ (t, xi) | 0leq t leq xi })=int_0^{infty}int_0^xi f(t,xi)dtdxi$$
However, I am confused how I can calculate $f(t, xi)$. I know from the question that $f(xi) = lambda e^{-lambda xi}$ where $lambda$ is the parameter of the geometric distribution. But I don't see how to get $f(t,xi)$
stochastic-processes
asked Nov 27 '18 at 8:58
AspiringMat
545518
A random process $X(t)$ is defined as $X(t)=1, 0leq t leq Y$ and $0$ otherwise where $Y$ follows an exponential distribution. What is the $pmf$ of $X(t)$
I am a bit confused on the definition of $pmf$ for random process. By definition, the pmf of $X(t)$ is
$$Pr({ X(t, xi)=1 })=Pr({ (t, xi) | 0leq t leq xi })=int_0^{infty}int_0^xi f(t,xi)dtdxi$$
However, I am confused how I can calculate $f(t, xi)$. I know from the question that $f(xi) = lambda e^{-lambda xi}$ where $lambda$ is the parameter of the geometric distribution. But I don't see how to get $f(t,xi)$
stochastic-processes
stochastic-processes
asked Nov 27 '18 at 8:58
AspiringMat
545518
asked Nov 27 '18 at 8:58
AspiringMat
545518
edited Nov 27 '18 at 9:09
asked Nov 27 '18 at 8:58
AspiringMat
545518
asked Nov 27 '18 at 8:58
AspiringMat
545518
asked Nov 27 '18 at 8:58
AspiringMat
545518
545518
My guess is that the question asks you to find the distribution of $X(t)$ for a fixed $t$ (and this is trivial). The distribution of the entire process is a measure on an infinite dimensional function space.
– Kavi Rama Murthy
Nov 27 '18 at 9:13
@KaviRamaMurthy That makes a lot more sense now. Just integrate from $t$ to infinity the geometric pdf. This is one of the reason I don't like the notation of suppressing the sample space in random processes. It just makes things more confusing for learners such as myself. Thank you!
– AspiringMat
Nov 27 '18 at 9:22
add a comment |
My guess is that the question asks you to find the distribution of $X(t)$ for a fixed $t$ (and this is trivial). The distribution of the entire process is a measure on an infinite dimensional function space.
– Kavi Rama Murthy
Nov 27 '18 at 9:13
@KaviRamaMurthy That makes a lot more sense now. Just integrate from $t$ to infinity the geometric pdf. This is one of the reason I don't like the notation of suppressing the sample space in random processes. It just makes things more confusing for learners such as myself. Thank you!
– AspiringMat
Nov 27 '18 at 9:22
My guess is that the question asks you to find the distribution of $X(t)$ for a fixed $t$ (and this is trivial). The distribution of the entire process is a measure on an infinite dimensional function space.
– Kavi Rama Murthy
Nov 27 '18 at 9:13
My guess is that the question asks you to find the distribution of $X(t)$ for a fixed $t$ (and this is trivial). The distribution of the entire process is a measure on an infinite dimensional function space.
– Kavi Rama Murthy
Nov 27 '18 at 9:13
@KaviRamaMurthy That makes a lot more sense now. Just integrate from $t$ to infinity the geometric pdf. This is one of the reason I don't like the notation of suppressing the sample space in random processes. It just makes things more confusing for learners such as myself. Thank you!
– AspiringMat
Nov 27 '18 at 9:22
@KaviRamaMurthy That makes a lot more sense now. Just integrate from $t$ to infinity the geometric pdf. This is one of the reason I don't like the notation of suppressing the sample space in random processes. It just makes things more confusing for learners such as myself. Thank you!
– AspiringMat
Nov 27 '18 at 9:22
add a comment |
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My guess is that the question asks you to find the distribution of $X(t)$ for a fixed $t$ (and this is trivial). The distribution of the entire process is a measure on an infinite dimensional function space.
– Kavi Rama Murthy
Nov 27 '18 at 9:13
@KaviRamaMurthy That makes a lot more sense now. Just integrate from $t$ to infinity the geometric pdf. This is one of the reason I don't like the notation of suppressing the sample space in random processes. It just makes things more confusing for learners such as myself. Thank you!
– AspiringMat
Nov 27 '18 at 9:22