Exponential pdf interpretation The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraProbability distribution for the age of a process prior to a stopping point (at an exponentially distributed time)Exponential Distribution with changing (time-varying) rate parameterExponential random Variable and Deterministic Variable MixWhen to assume exponential distribution?Given $f_Y(y;lambda) = lambda e^-lambda y$ , $y > 0 $, show $hat lambda = Y_1$ is not consistent for $lambda$exponential distribution question (poisson process)Given a variable $X$ with a PDF, what is the PDF of $sqrtX$Interarrival time in queu(e)-ing processinter arrival time ( continuous exponential distribution)How to get probability density functions (PDF) for an $N times D$ matrix?
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Exponential pdf interpretation
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraProbability distribution for the age of a process prior to a stopping point (at an exponentially distributed time)Exponential Distribution with changing (time-varying) rate parameterExponential random Variable and Deterministic Variable MixWhen to assume exponential distribution?Given $f_Y(y;lambda) = lambda e^-lambda y$ , $y > 0 $, show $hat lambda = Y_1$ is not consistent for $lambda$exponential distribution question (poisson process)Given a variable $X$ with a PDF, what is the PDF of $sqrtX$Interarrival time in queu(e)-ing processinter arrival time ( continuous exponential distribution)How to get probability density functions (PDF) for an $N times D$ matrix?
$begingroup$
I am looking at an exponential pdf in the context of interarrival times. The intensity is $2$ per unit of time. I plot the pdf of
$$f_Y_1 (y) = lambda e^-lambda y, y geq 0$$
What is $y$ here, it is time, right?
If I plot this pdf, I get:
Which does not make much sense.
Should not pdf give as a result the probability? Do I need to use scale constant, to turn the output into probability?
Is the output of the pdf an interarrival time instead of the probability?
I need to sample interarrival times from this pdf. How do I go about this?
probability
asked Mar 24 at 17:24
i squared - Keep it Reali squared - Keep it Real
1,63211128
$endgroup$
add a comment |
$begingroup$
I am looking at an exponential pdf in the context of interarrival times. The intensity is $2$ per unit of time. I plot the pdf of
$$f_Y_1 (y) = lambda e^-lambda y, y geq 0$$
What is $y$ here, it is time, right?
If I plot this pdf, I get:
Which does not make much sense.
Should not pdf give as a result the probability? Do I need to use scale constant, to turn the output into probability?
Is the output of the pdf an interarrival time instead of the probability?
I need to sample interarrival times from this pdf. How do I go about this?
probability
asked Mar 24 at 17:24
i squared - Keep it Reali squared - Keep it Real
1,63211128
$endgroup$
add a comment |
$begingroup$
I am looking at an exponential pdf in the context of interarrival times. The intensity is $2$ per unit of time. I plot the pdf of
$$f_Y_1 (y) = lambda e^-lambda y, y geq 0$$
What is $y$ here, it is time, right?
If I plot this pdf, I get:
Which does not make much sense.
Should not pdf give as a result the probability? Do I need to use scale constant, to turn the output into probability?
Is the output of the pdf an interarrival time instead of the probability?
I need to sample interarrival times from this pdf. How do I go about this?
probability
asked Mar 24 at 17:24
i squared - Keep it Reali squared - Keep it Real
1,63211128
$endgroup$
I am looking at an exponential pdf in the context of interarrival times. The intensity is $2$ per unit of time. I plot the pdf of
$$f_Y_1 (y) = lambda e^-lambda y, y geq 0$$
What is $y$ here, it is time, right?
If I plot this pdf, I get:
Which does not make much sense.
Should not pdf give as a result the probability? Do I need to use scale constant, to turn the output into probability?
Is the output of the pdf an interarrival time instead of the probability?
I need to sample interarrival times from this pdf. How do I go about this?
probability
probability
asked Mar 24 at 17:24
i squared - Keep it Reali squared - Keep it Real
1,63211128
asked Mar 24 at 17:24
i squared - Keep it Reali squared - Keep it Real
1,63211128
asked Mar 24 at 17:24
i squared - Keep it Reali squared - Keep it Real
1,63211128
asked Mar 24 at 17:24
i squared - Keep it Reali squared - Keep it Real
1,63211128
asked Mar 24 at 17:24
i squared - Keep it Reali squared - Keep it Real
1,63211128
1,63211128
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The software has computed $f$ at a small number of far apart points, turn joined them with lines. If you adjust the settings, you should get a smooth curve. The probability that $ale yle b$ is $int_a^b f(y) dy$. See here for how to sample $Y$.
answered Mar 24 at 17:27
J.G.J.G.
33.3k23252
$endgroup$
$begingroup$
I did that before (the smooth curve). It is just a smoother curve.
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:30
$begingroup$
I think $lambda$ that I should have used in my equation is not $2$, but $1/2$
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:37
$begingroup$
Well, with $2$ events per unit time the time between consecutive events has mean $1/lambda=1/2$, so yes, you need $lambda=2$. It's called a rate parameter.
$endgroup$
– J.G.
Mar 24 at 17:40
$begingroup$
k coo. So I can ignore the output from above. The probability is what you have defined it to be. And To sample I invert it, or just use the numpy's np.random.exponential :)
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:49
1
$begingroup$
Give scipy.stats.exponential a look as well. That way, it can do more than just sample (which you do with scipy.stats.exponential.rvs).
$endgroup$
– J.G.
Mar 24 at 17:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The software has computed $f$ at a small number of far apart points, turn joined them with lines. If you adjust the settings, you should get a smooth curve. The probability that $ale yle b$ is $int_a^b f(y) dy$. See here for how to sample $Y$.
answered Mar 24 at 17:27
J.G.J.G.
33.3k23252
$endgroup$
$begingroup$
I did that before (the smooth curve). It is just a smoother curve.
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:30
$begingroup$
I think $lambda$ that I should have used in my equation is not $2$, but $1/2$
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:37
$begingroup$
Well, with $2$ events per unit time the time between consecutive events has mean $1/lambda=1/2$, so yes, you need $lambda=2$. It's called a rate parameter.
$endgroup$
– J.G.
Mar 24 at 17:40
$begingroup$
k coo. So I can ignore the output from above. The probability is what you have defined it to be. And To sample I invert it, or just use the numpy's np.random.exponential :)
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:49
1
$begingroup$
Give scipy.stats.exponential a look as well. That way, it can do more than just sample (which you do with scipy.stats.exponential.rvs).
$endgroup$
– J.G.
Mar 24 at 17:57
add a comment |
$begingroup$
The software has computed $f$ at a small number of far apart points, turn joined them with lines. If you adjust the settings, you should get a smooth curve. The probability that $ale yle b$ is $int_a^b f(y) dy$. See here for how to sample $Y$.
answered Mar 24 at 17:27
J.G.J.G.
33.3k23252
$endgroup$
$begingroup$
I did that before (the smooth curve). It is just a smoother curve.
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:30
$begingroup$
I think $lambda$ that I should have used in my equation is not $2$, but $1/2$
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:37
$begingroup$
Well, with $2$ events per unit time the time between consecutive events has mean $1/lambda=1/2$, so yes, you need $lambda=2$. It's called a rate parameter.
$endgroup$
– J.G.
Mar 24 at 17:40
$begingroup$
k coo. So I can ignore the output from above. The probability is what you have defined it to be. And To sample I invert it, or just use the numpy's np.random.exponential :)
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:49
1
$begingroup$
Give scipy.stats.exponential a look as well. That way, it can do more than just sample (which you do with scipy.stats.exponential.rvs).
$endgroup$
– J.G.
Mar 24 at 17:57
add a comment |
$begingroup$
The software has computed $f$ at a small number of far apart points, turn joined them with lines. If you adjust the settings, you should get a smooth curve. The probability that $ale yle b$ is $int_a^b f(y) dy$. See here for how to sample $Y$.
answered Mar 24 at 17:27
J.G.J.G.
33.3k23252
$endgroup$
The software has computed $f$ at a small number of far apart points, turn joined them with lines. If you adjust the settings, you should get a smooth curve. The probability that $ale yle b$ is $int_a^b f(y) dy$. See here for how to sample $Y$.
answered Mar 24 at 17:27
J.G.J.G.
33.3k23252
answered Mar 24 at 17:27
J.G.J.G.
33.3k23252
answered Mar 24 at 17:27
J.G.J.G.
33.3k23252
answered Mar 24 at 17:27
J.G.J.G.
33.3k23252
33.3k23252
$begingroup$
I did that before (the smooth curve). It is just a smoother curve.
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:30
$begingroup$
I think $lambda$ that I should have used in my equation is not $2$, but $1/2$
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:37
$begingroup$
Well, with $2$ events per unit time the time between consecutive events has mean $1/lambda=1/2$, so yes, you need $lambda=2$. It's called a rate parameter.
$endgroup$
– J.G.
Mar 24 at 17:40
$begingroup$
k coo. So I can ignore the output from above. The probability is what you have defined it to be. And To sample I invert it, or just use the numpy's np.random.exponential :)
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:49
1
$begingroup$
Give scipy.stats.exponential a look as well. That way, it can do more than just sample (which you do with scipy.stats.exponential.rvs).
$endgroup$
– J.G.
Mar 24 at 17:57
add a comment |
$begingroup$
I did that before (the smooth curve). It is just a smoother curve.
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:30
$begingroup$
I think $lambda$ that I should have used in my equation is not $2$, but $1/2$
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:37
$begingroup$
Well, with $2$ events per unit time the time between consecutive events has mean $1/lambda=1/2$, so yes, you need $lambda=2$. It's called a rate parameter.
$endgroup$
– J.G.
Mar 24 at 17:40
$begingroup$
k coo. So I can ignore the output from above. The probability is what you have defined it to be. And To sample I invert it, or just use the numpy's np.random.exponential :)
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:49
1
$begingroup$
Give scipy.stats.exponential a look as well. That way, it can do more than just sample (which you do with scipy.stats.exponential.rvs).
$endgroup$
– J.G.
Mar 24 at 17:57
$begingroup$
I did that before (the smooth curve). It is just a smoother curve.
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:30
$begingroup$
I did that before (the smooth curve). It is just a smoother curve.
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:30
$begingroup$
I think $lambda$ that I should have used in my equation is not $2$, but $1/2$
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:37
$begingroup$
I think $lambda$ that I should have used in my equation is not $2$, but $1/2$
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:37
$begingroup$
Well, with $2$ events per unit time the time between consecutive events has mean $1/lambda=1/2$, so yes, you need $lambda=2$. It's called a rate parameter.
$endgroup$
– J.G.
Mar 24 at 17:40
$begingroup$
Well, with $2$ events per unit time the time between consecutive events has mean $1/lambda=1/2$, so yes, you need $lambda=2$. It's called a rate parameter.
$endgroup$
– J.G.
Mar 24 at 17:40
$begingroup$
k coo. So I can ignore the output from above. The probability is what you have defined it to be. And To sample I invert it, or just use the numpy's np.random.exponential :)
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:49
$begingroup$
k coo. So I can ignore the output from above. The probability is what you have defined it to be. And To sample I invert it, or just use the numpy's np.random.exponential :)
$endgroup$
– i squared - Keep it Real
Mar 24 at 17:49
1
1
$begingroup$
Give scipy.stats.exponential a look as well. That way, it can do more than just sample (which you do with scipy.stats.exponential.rvs).
$endgroup$
– J.G.
Mar 24 at 17:57
$begingroup$
Give scipy.stats.exponential a look as well. That way, it can do more than just sample (which you do with scipy.stats.exponential.rvs).
$endgroup$
– J.G.
Mar 24 at 17:57
add a comment |
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