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Notation of marginal probabilities
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I found a (to me) strange notation concerning marginal probabilities I don’t understand. Unfortunately I will include picture of the notation.
Does it mean x=2.5, shouldn’t it be 0 then? Do they mean the cumulative distribution fuchtion?
Thanks
marginal-probability
edited Jul 14 at 23:50
Arnaud Mortier
19.2k22159
asked Jul 14 at 23:37
Lillys
84
add a comment |Â
up vote
0
down vote
favorite
I found a (to me) strange notation concerning marginal probabilities I don’t understand. Unfortunately I will include picture of the notation.
Does it mean x=2.5, shouldn’t it be 0 then? Do they mean the cumulative distribution fuchtion?
Thanks
marginal-probability
edited Jul 14 at 23:50
Arnaud Mortier
19.2k22159
asked Jul 14 at 23:37
Lillys
84
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I found a (to me) strange notation concerning marginal probabilities I don’t understand. Unfortunately I will include picture of the notation.
Does it mean x=2.5, shouldn’t it be 0 then? Do they mean the cumulative distribution fuchtion?
Thanks
marginal-probability
edited Jul 14 at 23:50
Arnaud Mortier
19.2k22159
asked Jul 14 at 23:37
Lillys
84
I found a (to me) strange notation concerning marginal probabilities I don’t understand. Unfortunately I will include picture of the notation.
Does it mean x=2.5, shouldn’t it be 0 then? Do they mean the cumulative distribution fuchtion?
Thanks
marginal-probability
edited Jul 14 at 23:50
Arnaud Mortier
19.2k22159
asked Jul 14 at 23:37
Lillys
84
edited Jul 14 at 23:50
Arnaud Mortier
19.2k22159
edited Jul 14 at 23:50
Arnaud Mortier
19.2k22159
edited Jul 14 at 23:50
Arnaud Mortier
19.2k22159
19.2k22159
asked Jul 14 at 23:37
Lillys
84
asked Jul 14 at 23:37
Lillys
84
asked Jul 14 at 23:37
Lillys
84
84
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
They mean the density, which is equal to the derivative of the CDF wherever this one is differentiable.
The notation for a CDF would be $F_X(x)$ (with a capital $F$).
Both are linked by the fundamental $$F_X(x)=int_-infty^x f_X(t)dt$$
Note that such a density does not necessarily exist. When it does, $X$ is called absolutely continuous.
Note also that $f_X$ has little to do with the probability mass function (as you seem to believe when you say that it has to be $0$), which is denoted by $p_X$.
answered Jul 14 at 23:47
Arnaud Mortier
19.2k22159
@Lillys $f_X$ is a function, so it can be evaluated at a point. No, it's not necessarily $0$. It is not a PMF. For instance an exponential RV has CDF $1-e^-lambda x$ and PDF $lambda e^-lambda x$. Replace $x$ by $2.5$ to get the value of the PDF at $2.5$.
– Arnaud Mortier
Jul 14 at 23:51
than a value of 2.5 would be the probability that x takes on a value of 2.5? I understand that a function can be evaluated at that point. But what does that mean in terms of probability.
– Lillys
Jul 14 at 23:55
The pmf exists for all distributions, the only thing is that for continuous RVs it does not carry any information so you never talk about it. But there are mixed-type RVs as well, not continuous, but not discrete, where the PMF does carry some incomplete information.
– Arnaud Mortier
Jul 14 at 23:57
Thanks for your answer !
– Lillys
Jul 15 at 0:00
@Lillys You're welcome. I see that you edited your comment. Note that $f_X(x)$ is a density of probability, which is different from a probability.
– Arnaud Mortier
Jul 15 at 0:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
They mean the density, which is equal to the derivative of the CDF wherever this one is differentiable.
The notation for a CDF would be $F_X(x)$ (with a capital $F$).
Both are linked by the fundamental $$F_X(x)=int_-infty^x f_X(t)dt$$
Note that such a density does not necessarily exist. When it does, $X$ is called absolutely continuous.
Note also that $f_X$ has little to do with the probability mass function (as you seem to believe when you say that it has to be $0$), which is denoted by $p_X$.
answered Jul 14 at 23:47
Arnaud Mortier
19.2k22159
@Lillys $f_X$ is a function, so it can be evaluated at a point. No, it's not necessarily $0$. It is not a PMF. For instance an exponential RV has CDF $1-e^-lambda x$ and PDF $lambda e^-lambda x$. Replace $x$ by $2.5$ to get the value of the PDF at $2.5$.
– Arnaud Mortier
Jul 14 at 23:51
than a value of 2.5 would be the probability that x takes on a value of 2.5? I understand that a function can be evaluated at that point. But what does that mean in terms of probability.
– Lillys
Jul 14 at 23:55
The pmf exists for all distributions, the only thing is that for continuous RVs it does not carry any information so you never talk about it. But there are mixed-type RVs as well, not continuous, but not discrete, where the PMF does carry some incomplete information.
– Arnaud Mortier
Jul 14 at 23:57
Thanks for your answer !
– Lillys
Jul 15 at 0:00
@Lillys You're welcome. I see that you edited your comment. Note that $f_X(x)$ is a density of probability, which is different from a probability.
– Arnaud Mortier
Jul 15 at 0:02
add a comment |Â
up vote
0
down vote
accepted
They mean the density, which is equal to the derivative of the CDF wherever this one is differentiable.
The notation for a CDF would be $F_X(x)$ (with a capital $F$).
Both are linked by the fundamental $$F_X(x)=int_-infty^x f_X(t)dt$$
Note that such a density does not necessarily exist. When it does, $X$ is called absolutely continuous.
Note also that $f_X$ has little to do with the probability mass function (as you seem to believe when you say that it has to be $0$), which is denoted by $p_X$.
answered Jul 14 at 23:47
Arnaud Mortier
19.2k22159
@Lillys $f_X$ is a function, so it can be evaluated at a point. No, it's not necessarily $0$. It is not a PMF. For instance an exponential RV has CDF $1-e^-lambda x$ and PDF $lambda e^-lambda x$. Replace $x$ by $2.5$ to get the value of the PDF at $2.5$.
– Arnaud Mortier
Jul 14 at 23:51
than a value of 2.5 would be the probability that x takes on a value of 2.5? I understand that a function can be evaluated at that point. But what does that mean in terms of probability.
– Lillys
Jul 14 at 23:55
The pmf exists for all distributions, the only thing is that for continuous RVs it does not carry any information so you never talk about it. But there are mixed-type RVs as well, not continuous, but not discrete, where the PMF does carry some incomplete information.
– Arnaud Mortier
Jul 14 at 23:57
Thanks for your answer !
– Lillys
Jul 15 at 0:00
@Lillys You're welcome. I see that you edited your comment. Note that $f_X(x)$ is a density of probability, which is different from a probability.
– Arnaud Mortier
Jul 15 at 0:02
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
They mean the density, which is equal to the derivative of the CDF wherever this one is differentiable.
The notation for a CDF would be $F_X(x)$ (with a capital $F$).
Both are linked by the fundamental $$F_X(x)=int_-infty^x f_X(t)dt$$
Note that such a density does not necessarily exist. When it does, $X$ is called absolutely continuous.
Note also that $f_X$ has little to do with the probability mass function (as you seem to believe when you say that it has to be $0$), which is denoted by $p_X$.
answered Jul 14 at 23:47
Arnaud Mortier
19.2k22159
They mean the density, which is equal to the derivative of the CDF wherever this one is differentiable.
The notation for a CDF would be $F_X(x)$ (with a capital $F$).
Both are linked by the fundamental $$F_X(x)=int_-infty^x f_X(t)dt$$
Note that such a density does not necessarily exist. When it does, $X$ is called absolutely continuous.
Note also that $f_X$ has little to do with the probability mass function (as you seem to believe when you say that it has to be $0$), which is denoted by $p_X$.
answered Jul 14 at 23:47
Arnaud Mortier
19.2k22159
answered Jul 14 at 23:47
Arnaud Mortier
19.2k22159
answered Jul 14 at 23:47
Arnaud Mortier
19.2k22159
answered Jul 14 at 23:47
Arnaud Mortier
19.2k22159
19.2k22159
@Lillys $f_X$ is a function, so it can be evaluated at a point. No, it's not necessarily $0$. It is not a PMF. For instance an exponential RV has CDF $1-e^-lambda x$ and PDF $lambda e^-lambda x$. Replace $x$ by $2.5$ to get the value of the PDF at $2.5$.
– Arnaud Mortier
Jul 14 at 23:51
than a value of 2.5 would be the probability that x takes on a value of 2.5? I understand that a function can be evaluated at that point. But what does that mean in terms of probability.
– Lillys
Jul 14 at 23:55
The pmf exists for all distributions, the only thing is that for continuous RVs it does not carry any information so you never talk about it. But there are mixed-type RVs as well, not continuous, but not discrete, where the PMF does carry some incomplete information.
– Arnaud Mortier
Jul 14 at 23:57
Thanks for your answer !
– Lillys
Jul 15 at 0:00
@Lillys You're welcome. I see that you edited your comment. Note that $f_X(x)$ is a density of probability, which is different from a probability.
– Arnaud Mortier
Jul 15 at 0:02
add a comment |Â
@Lillys $f_X$ is a function, so it can be evaluated at a point. No, it's not necessarily $0$. It is not a PMF. For instance an exponential RV has CDF $1-e^-lambda x$ and PDF $lambda e^-lambda x$. Replace $x$ by $2.5$ to get the value of the PDF at $2.5$.
– Arnaud Mortier
Jul 14 at 23:51
than a value of 2.5 would be the probability that x takes on a value of 2.5? I understand that a function can be evaluated at that point. But what does that mean in terms of probability.
– Lillys
Jul 14 at 23:55
The pmf exists for all distributions, the only thing is that for continuous RVs it does not carry any information so you never talk about it. But there are mixed-type RVs as well, not continuous, but not discrete, where the PMF does carry some incomplete information.
– Arnaud Mortier
Jul 14 at 23:57
Thanks for your answer !
– Lillys
Jul 15 at 0:00
@Lillys You're welcome. I see that you edited your comment. Note that $f_X(x)$ is a density of probability, which is different from a probability.
– Arnaud Mortier
Jul 15 at 0:02
@Lillys $f_X$ is a function, so it can be evaluated at a point. No, it's not necessarily $0$. It is not a PMF. For instance an exponential RV has CDF $1-e^-lambda x$ and PDF $lambda e^-lambda x$. Replace $x$ by $2.5$ to get the value of the PDF at $2.5$.
– Arnaud Mortier
Jul 14 at 23:51
@Lillys $f_X$ is a function, so it can be evaluated at a point. No, it's not necessarily $0$. It is not a PMF. For instance an exponential RV has CDF $1-e^-lambda x$ and PDF $lambda e^-lambda x$. Replace $x$ by $2.5$ to get the value of the PDF at $2.5$.
– Arnaud Mortier
Jul 14 at 23:51
than a value of 2.5 would be the probability that x takes on a value of 2.5? I understand that a function can be evaluated at that point. But what does that mean in terms of probability.
– Lillys
Jul 14 at 23:55
than a value of 2.5 would be the probability that x takes on a value of 2.5? I understand that a function can be evaluated at that point. But what does that mean in terms of probability.
– Lillys
Jul 14 at 23:55
The pmf exists for all distributions, the only thing is that for continuous RVs it does not carry any information so you never talk about it. But there are mixed-type RVs as well, not continuous, but not discrete, where the PMF does carry some incomplete information.
– Arnaud Mortier
Jul 14 at 23:57
The pmf exists for all distributions, the only thing is that for continuous RVs it does not carry any information so you never talk about it. But there are mixed-type RVs as well, not continuous, but not discrete, where the PMF does carry some incomplete information.
– Arnaud Mortier
Jul 14 at 23:57
Thanks for your answer !
– Lillys
Jul 15 at 0:00
Thanks for your answer !
– Lillys
Jul 15 at 0:00
@Lillys You're welcome. I see that you edited your comment. Note that $f_X(x)$ is a density of probability, which is different from a probability.
– Arnaud Mortier
Jul 15 at 0:02
@Lillys You're welcome. I see that you edited your comment. Note that $f_X(x)$ is a density of probability, which is different from a probability.
– Arnaud Mortier
Jul 15 at 0:02
add a comment |Â
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