Mixing
What does the value of a probability density function (PDF) at some x indicate?
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I understand that the probability mass function of a discrete random-variable X is $y=g(x)$. This means $P(X=x_0) = g(x_0)$.
Now, a probability density function of of a continuous random variable X is $y=f(x)$. Wikipedia defines this function $y$ to mean
In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value.
I am confused about the meaning of 'relative likelihood' because it certainly does not mean probability! The probability $P(X<x_0)$ is given by some integral of the pdf.
So what does $f(x_0)$ indicate? It gives a real number, but isn't the relative likelihood of a specific value for a CRV always zero?
probability statistics random-variables
asked Oct 10 '12 at 19:20
jesterIIjesterII
1,20121326
$endgroup$
add a comment |
$begingroup$
I understand that the probability mass function of a discrete random-variable X is $y=g(x)$. This means $P(X=x_0) = g(x_0)$.
Now, a probability density function of of a continuous random variable X is $y=f(x)$. Wikipedia defines this function $y$ to mean
In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value.
I am confused about the meaning of 'relative likelihood' because it certainly does not mean probability! The probability $P(X<x_0)$ is given by some integral of the pdf.
So what does $f(x_0)$ indicate? It gives a real number, but isn't the relative likelihood of a specific value for a CRV always zero?
probability statistics random-variables
asked Oct 10 '12 at 19:20
jesterIIjesterII
1,20121326
$endgroup$
5
$begingroup$
Let $f$ be the density function of $X$. Assume $f$ is continuous. Then if $h$ is small, the probability that $X$ lies in the interval $[a,a+h]$ is approximately $hf(a)$. By approximately I mean that the probability, divided by $h$, approaches $f(a)$ as $h$ approaches $0$. So the ratio $f(a)/f(b)$ measures, approximately, the ratio of the probability that $X$ is in $[a,a+h]$ to the probability $X$ is in $[b,b+h]$.
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– André Nicolas
Oct 10 '12 at 19:28
add a comment |
$begingroup$
I understand that the probability mass function of a discrete random-variable X is $y=g(x)$. This means $P(X=x_0) = g(x_0)$.
Now, a probability density function of of a continuous random variable X is $y=f(x)$. Wikipedia defines this function $y$ to mean
In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value.
I am confused about the meaning of 'relative likelihood' because it certainly does not mean probability! The probability $P(X<x_0)$ is given by some integral of the pdf.
So what does $f(x_0)$ indicate? It gives a real number, but isn't the relative likelihood of a specific value for a CRV always zero?
probability statistics random-variables
asked Oct 10 '12 at 19:20
jesterIIjesterII
1,20121326
$endgroup$
I understand that the probability mass function of a discrete random-variable X is $y=g(x)$. This means $P(X=x_0) = g(x_0)$.
Now, a probability density function of of a continuous random variable X is $y=f(x)$. Wikipedia defines this function $y$ to mean
In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value.
I am confused about the meaning of 'relative likelihood' because it certainly does not mean probability! The probability $P(X<x_0)$ is given by some integral of the pdf.
So what does $f(x_0)$ indicate? It gives a real number, but isn't the relative likelihood of a specific value for a CRV always zero?
probability statistics random-variables
probability statistics random-variables
asked Oct 10 '12 at 19:20
jesterIIjesterII
1,20121326
asked Oct 10 '12 at 19:20
jesterIIjesterII
1,20121326
asked Oct 10 '12 at 19:20
jesterIIjesterII
1,20121326
asked Oct 10 '12 at 19:20
jesterIIjesterII
1,20121326
asked Oct 10 '12 at 19:20
jesterIIjesterII
1,20121326
1,20121326
5
$begingroup$
Let $f$ be the density function of $X$. Assume $f$ is continuous. Then if $h$ is small, the probability that $X$ lies in the interval $[a,a+h]$ is approximately $hf(a)$. By approximately I mean that the probability, divided by $h$, approaches $f(a)$ as $h$ approaches $0$. So the ratio $f(a)/f(b)$ measures, approximately, the ratio of the probability that $X$ is in $[a,a+h]$ to the probability $X$ is in $[b,b+h]$.
$endgroup$
– André Nicolas
Oct 10 '12 at 19:28
add a comment |
5
$begingroup$
Let $f$ be the density function of $X$. Assume $f$ is continuous. Then if $h$ is small, the probability that $X$ lies in the interval $[a,a+h]$ is approximately $hf(a)$. By approximately I mean that the probability, divided by $h$, approaches $f(a)$ as $h$ approaches $0$. So the ratio $f(a)/f(b)$ measures, approximately, the ratio of the probability that $X$ is in $[a,a+h]$ to the probability $X$ is in $[b,b+h]$.
$endgroup$
– André Nicolas
Oct 10 '12 at 19:28
5
5
$begingroup$
Let $f$ be the density function of $X$. Assume $f$ is continuous. Then if $h$ is small, the probability that $X$ lies in the interval $[a,a+h]$ is approximately $hf(a)$. By approximately I mean that the probability, divided by $h$, approaches $f(a)$ as $h$ approaches $0$. So the ratio $f(a)/f(b)$ measures, approximately, the ratio of the probability that $X$ is in $[a,a+h]$ to the probability $X$ is in $[b,b+h]$.
$endgroup$
– André Nicolas
Oct 10 '12 at 19:28
$begingroup$
Let $f$ be the density function of $X$. Assume $f$ is continuous. Then if $h$ is small, the probability that $X$ lies in the interval $[a,a+h]$ is approximately $hf(a)$. By approximately I mean that the probability, divided by $h$, approaches $f(a)$ as $h$ approaches $0$. So the ratio $f(a)/f(b)$ measures, approximately, the ratio of the probability that $X$ is in $[a,a+h]$ to the probability $X$ is in $[b,b+h]$.
$endgroup$
– André Nicolas
Oct 10 '12 at 19:28
add a comment |
5 Answers
5
active
oldest
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$begingroup$
'Relative likelihood' is indeed misleading. Look at it as a limit instead:
$$
f(x)=lim_{h to 0}frac{F(x+h)-F(x)}{h}
$$
where $F(x) = P(X leq x)$
answered Oct 10 '12 at 19:28
AlexAlex
14.2k42134
$endgroup$
$begingroup$
So you suggest looking at the pdf as being defined by the cumulative distribution function?
$endgroup$
– jesterII
Oct 10 '12 at 19:39
2
$begingroup$
This is essentially the definition of pdf fro CRVs
$endgroup$
– Alex
Oct 10 '12 at 20:12
2
$begingroup$
A good way of thinking about is $f(x) = frac{dF}{dx}$ and so it's the rate of change of the cdf at $x$.
$endgroup$
– Jacob
Feb 27 '13 at 17:39
$begingroup$
Hi Alex, sorry for asking problem related to such an old answer. I know the pdf $f$ is the derivative of the cdf $F$, but what is the physical meaning of "the rate of change of the cdf at some point"? I mean, how to explain it by using the continuous random variable $X$?
$endgroup$
– Sam Wong
Dec 13 '18 at 8:17
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I don't know much about physics sorry.
$endgroup$
– Alex
Dec 13 '18 at 10:55
add a comment |
$begingroup$
In general, if $X$ is a random variable with values of a measure space $(A,mathcal A,mu)$ and with pdf $f:Ato [0,1]$, then for all measurable set $Sinmathcal A$,
$$P(Xin S) = int_S fdmu $$
So, if $A=Bbb R$ (and $mu=lambda$), then
$$P(a<X<b)=int_a^b f(x)dx$$
So, $f(x) = displaystylelim_{tto 0} frac1{2t}int_{x-t}^{x+t} f =lim_{tto 0} frac1{2t} P(|X-x|<t) $ for example.. We can call it 'relative likelihood'..
answered Oct 10 '12 at 19:30
BerciBerci
60k23672
$endgroup$
1
$begingroup$
This is a better answer than Alex's but doesn't explain the significance of the number $f(x)$. Does it have a meaning independent of a cdf? Andre's answer of it being approximately $hf(a)$ is great but he doesn't indicate if there's more to $f(x)$ by itself.
$endgroup$
– Jacob
Feb 27 '13 at 17:23
add a comment |
$begingroup$
Intro statistics focuses on the PDF as the description of the population, but in fact it is the CDF (cumulative density function) that gives you a functional understanding of the population, as points on the CDF denote probabilities over a relevant range of measures. If you look at all stats from this perspective, then the PDF is just the description of probability change with respect to a change around a point along the measure at hand. The values on the PDF therefore only give you a look at the spread. For example, given two normal distributions $N(mu_1, sigma_1^2)$ and $N(mu_2, sigma_2^2)$, if you choose any value of $x$ to get point $p_n=mu_n+xcdotsigma_n$ for the respective distributions and get $X_1[p_1 ] > X_2[p_2 ]$, then this just means $sigma_1 < sigma_2$. Similar relationships exist for other distributions.
answered Jun 28 '15 at 14:44
TopherTopher
347317
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Very interesting answer!
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– information_interchange
Jun 7 '18 at 18:38
add a comment |
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I am not sure if Jester is still interested, as it's been 5 years, but I think I found a less confusing anwer than in Wikipedia.
In contrast to discrete random variables, if X is continuous, f(X) is a function whose value at any given sample is not the probability but rather it indicates the likelihood that X will be in that sample/interval. For example if the value of the PDF around a point (can be generalized for a sample) x is large, that means the random variable X is more likely to take values close to x. If, on the other hand, f(x)=0 in some interval, then X won't be in that interval
Of course a more practical way of thinking it is that the probability of X being in an interval is given by the integral of the PDF.
You might want to look at the link below for more details: http://mathinsight.org/probability_density_function_idea
answered Oct 24 '17 at 23:25
ALEX.VAMVASALEX.VAMVAS
211
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add a comment |
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The ratio of the pdf $f(x)$ at two points, $r_x = f(x_0)/f(x_1)$, is not a measure of relative probability (or "relative likelihood") for the two outcomes for the random variable $X$. The ratio depends on the metric. That is, with a variable transformation, $z=z(x)$, with the pdf for $Z$ given by $h(z)$, the ratio $r_z=h(z_0)/h(z_1)neq r_x$, in general, even though the two ratios refer to the same two outcomes. For monotonic transformations, $f(x),dx = h(z),dz$.
Numerical values of the pdf have no value on their own. The metric, $dx$, is required for probability interpretations (ie. $f(x),dx$). Wikipedia got this wrong, so I have corrected it.
answered Nov 27 '16 at 0:47
DeanDean
1,00537
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
'Relative likelihood' is indeed misleading. Look at it as a limit instead:
$$
f(x)=lim_{h to 0}frac{F(x+h)-F(x)}{h}
$$
where $F(x) = P(X leq x)$
answered Oct 10 '12 at 19:28
AlexAlex
14.2k42134
$endgroup$
$begingroup$
So you suggest looking at the pdf as being defined by the cumulative distribution function?
$endgroup$
– jesterII
Oct 10 '12 at 19:39
2
$begingroup$
This is essentially the definition of pdf fro CRVs
$endgroup$
– Alex
Oct 10 '12 at 20:12
2
$begingroup$
A good way of thinking about is $f(x) = frac{dF}{dx}$ and so it's the rate of change of the cdf at $x$.
$endgroup$
– Jacob
Feb 27 '13 at 17:39
$begingroup$
Hi Alex, sorry for asking problem related to such an old answer. I know the pdf $f$ is the derivative of the cdf $F$, but what is the physical meaning of "the rate of change of the cdf at some point"? I mean, how to explain it by using the continuous random variable $X$?
$endgroup$
– Sam Wong
Dec 13 '18 at 8:17
$begingroup$
I don't know much about physics sorry.
$endgroup$
– Alex
Dec 13 '18 at 10:55
add a comment |
$begingroup$
'Relative likelihood' is indeed misleading. Look at it as a limit instead:
$$
f(x)=lim_{h to 0}frac{F(x+h)-F(x)}{h}
$$
where $F(x) = P(X leq x)$
answered Oct 10 '12 at 19:28
AlexAlex
14.2k42134
$endgroup$
$begingroup$
So you suggest looking at the pdf as being defined by the cumulative distribution function?
$endgroup$
– jesterII
Oct 10 '12 at 19:39
2
$begingroup$
This is essentially the definition of pdf fro CRVs
$endgroup$
– Alex
Oct 10 '12 at 20:12
2
$begingroup$
A good way of thinking about is $f(x) = frac{dF}{dx}$ and so it's the rate of change of the cdf at $x$.
$endgroup$
– Jacob
Feb 27 '13 at 17:39
$begingroup$
Hi Alex, sorry for asking problem related to such an old answer. I know the pdf $f$ is the derivative of the cdf $F$, but what is the physical meaning of "the rate of change of the cdf at some point"? I mean, how to explain it by using the continuous random variable $X$?
$endgroup$
– Sam Wong
Dec 13 '18 at 8:17
$begingroup$
I don't know much about physics sorry.
$endgroup$
– Alex
Dec 13 '18 at 10:55
add a comment |
$begingroup$
'Relative likelihood' is indeed misleading. Look at it as a limit instead:
$$
f(x)=lim_{h to 0}frac{F(x+h)-F(x)}{h}
$$
where $F(x) = P(X leq x)$
answered Oct 10 '12 at 19:28
AlexAlex
14.2k42134
$endgroup$
'Relative likelihood' is indeed misleading. Look at it as a limit instead:
$$
f(x)=lim_{h to 0}frac{F(x+h)-F(x)}{h}
$$
where $F(x) = P(X leq x)$
answered Oct 10 '12 at 19:28
AlexAlex
14.2k42134
answered Oct 10 '12 at 19:28
AlexAlex
14.2k42134
answered Oct 10 '12 at 19:28
AlexAlex
14.2k42134
answered Oct 10 '12 at 19:28
AlexAlex
14.2k42134
14.2k42134
$begingroup$
So you suggest looking at the pdf as being defined by the cumulative distribution function?
$endgroup$
– jesterII
Oct 10 '12 at 19:39
2
$begingroup$
This is essentially the definition of pdf fro CRVs
$endgroup$
– Alex
Oct 10 '12 at 20:12
2
$begingroup$
A good way of thinking about is $f(x) = frac{dF}{dx}$ and so it's the rate of change of the cdf at $x$.
$endgroup$
– Jacob
Feb 27 '13 at 17:39
$begingroup$
Hi Alex, sorry for asking problem related to such an old answer. I know the pdf $f$ is the derivative of the cdf $F$, but what is the physical meaning of "the rate of change of the cdf at some point"? I mean, how to explain it by using the continuous random variable $X$?
$endgroup$
– Sam Wong
Dec 13 '18 at 8:17
$begingroup$
I don't know much about physics sorry.
$endgroup$
– Alex
Dec 13 '18 at 10:55
add a comment |
$begingroup$
So you suggest looking at the pdf as being defined by the cumulative distribution function?
$endgroup$
– jesterII
Oct 10 '12 at 19:39
2
$begingroup$
This is essentially the definition of pdf fro CRVs
$endgroup$
– Alex
Oct 10 '12 at 20:12
2
$begingroup$
A good way of thinking about is $f(x) = frac{dF}{dx}$ and so it's the rate of change of the cdf at $x$.
$endgroup$
– Jacob
Feb 27 '13 at 17:39
$begingroup$
Hi Alex, sorry for asking problem related to such an old answer. I know the pdf $f$ is the derivative of the cdf $F$, but what is the physical meaning of "the rate of change of the cdf at some point"? I mean, how to explain it by using the continuous random variable $X$?
$endgroup$
– Sam Wong
Dec 13 '18 at 8:17
$begingroup$
I don't know much about physics sorry.
$endgroup$
– Alex
Dec 13 '18 at 10:55
$begingroup$
So you suggest looking at the pdf as being defined by the cumulative distribution function?
$endgroup$
– jesterII
Oct 10 '12 at 19:39
$begingroup$
So you suggest looking at the pdf as being defined by the cumulative distribution function?
$endgroup$
– jesterII
Oct 10 '12 at 19:39
2
2
$begingroup$
This is essentially the definition of pdf fro CRVs
$endgroup$
– Alex
Oct 10 '12 at 20:12
$begingroup$
This is essentially the definition of pdf fro CRVs
$endgroup$
– Alex
Oct 10 '12 at 20:12
2
2
$begingroup$
A good way of thinking about is $f(x) = frac{dF}{dx}$ and so it's the rate of change of the cdf at $x$.
$endgroup$
– Jacob
Feb 27 '13 at 17:39
$begingroup$
A good way of thinking about is $f(x) = frac{dF}{dx}$ and so it's the rate of change of the cdf at $x$.
$endgroup$
– Jacob
Feb 27 '13 at 17:39
$begingroup$
Hi Alex, sorry for asking problem related to such an old answer. I know the pdf $f$ is the derivative of the cdf $F$, but what is the physical meaning of "the rate of change of the cdf at some point"? I mean, how to explain it by using the continuous random variable $X$?
$endgroup$
– Sam Wong
Dec 13 '18 at 8:17
$begingroup$
Hi Alex, sorry for asking problem related to such an old answer. I know the pdf $f$ is the derivative of the cdf $F$, but what is the physical meaning of "the rate of change of the cdf at some point"? I mean, how to explain it by using the continuous random variable $X$?
$endgroup$
– Sam Wong
Dec 13 '18 at 8:17
$begingroup$
I don't know much about physics sorry.
$endgroup$
– Alex
Dec 13 '18 at 10:55
$begingroup$
I don't know much about physics sorry.
$endgroup$
– Alex
Dec 13 '18 at 10:55
add a comment |
$begingroup$
In general, if $X$ is a random variable with values of a measure space $(A,mathcal A,mu)$ and with pdf $f:Ato [0,1]$, then for all measurable set $Sinmathcal A$,
$$P(Xin S) = int_S fdmu $$
So, if $A=Bbb R$ (and $mu=lambda$), then
$$P(a<X<b)=int_a^b f(x)dx$$
So, $f(x) = displaystylelim_{tto 0} frac1{2t}int_{x-t}^{x+t} f =lim_{tto 0} frac1{2t} P(|X-x|<t) $ for example.. We can call it 'relative likelihood'..
answered Oct 10 '12 at 19:30
BerciBerci
60k23672
$endgroup$
1
$begingroup$
This is a better answer than Alex's but doesn't explain the significance of the number $f(x)$. Does it have a meaning independent of a cdf? Andre's answer of it being approximately $hf(a)$ is great but he doesn't indicate if there's more to $f(x)$ by itself.
$endgroup$
– Jacob
Feb 27 '13 at 17:23
add a comment |
$begingroup$
In general, if $X$ is a random variable with values of a measure space $(A,mathcal A,mu)$ and with pdf $f:Ato [0,1]$, then for all measurable set $Sinmathcal A$,
$$P(Xin S) = int_S fdmu $$
So, if $A=Bbb R$ (and $mu=lambda$), then
$$P(a<X<b)=int_a^b f(x)dx$$
So, $f(x) = displaystylelim_{tto 0} frac1{2t}int_{x-t}^{x+t} f =lim_{tto 0} frac1{2t} P(|X-x|<t) $ for example.. We can call it 'relative likelihood'..
answered Oct 10 '12 at 19:30
BerciBerci
60k23672
$endgroup$
1
$begingroup$
This is a better answer than Alex's but doesn't explain the significance of the number $f(x)$. Does it have a meaning independent of a cdf? Andre's answer of it being approximately $hf(a)$ is great but he doesn't indicate if there's more to $f(x)$ by itself.
$endgroup$
– Jacob
Feb 27 '13 at 17:23
add a comment |
$begingroup$
In general, if $X$ is a random variable with values of a measure space $(A,mathcal A,mu)$ and with pdf $f:Ato [0,1]$, then for all measurable set $Sinmathcal A$,
$$P(Xin S) = int_S fdmu $$
So, if $A=Bbb R$ (and $mu=lambda$), then
$$P(a<X<b)=int_a^b f(x)dx$$
So, $f(x) = displaystylelim_{tto 0} frac1{2t}int_{x-t}^{x+t} f =lim_{tto 0} frac1{2t} P(|X-x|<t) $ for example.. We can call it 'relative likelihood'..
answered Oct 10 '12 at 19:30
BerciBerci
60k23672
$endgroup$
In general, if $X$ is a random variable with values of a measure space $(A,mathcal A,mu)$ and with pdf $f:Ato [0,1]$, then for all measurable set $Sinmathcal A$,
$$P(Xin S) = int_S fdmu $$
So, if $A=Bbb R$ (and $mu=lambda$), then
$$P(a<X<b)=int_a^b f(x)dx$$
So, $f(x) = displaystylelim_{tto 0} frac1{2t}int_{x-t}^{x+t} f =lim_{tto 0} frac1{2t} P(|X-x|<t) $ for example.. We can call it 'relative likelihood'..
answered Oct 10 '12 at 19:30
BerciBerci
60k23672
answered Oct 10 '12 at 19:30
BerciBerci
60k23672
answered Oct 10 '12 at 19:30
BerciBerci
60k23672
answered Oct 10 '12 at 19:30
BerciBerci
60k23672
60k23672
1
$begingroup$
This is a better answer than Alex's but doesn't explain the significance of the number $f(x)$. Does it have a meaning independent of a cdf? Andre's answer of it being approximately $hf(a)$ is great but he doesn't indicate if there's more to $f(x)$ by itself.
$endgroup$
– Jacob
Feb 27 '13 at 17:23
add a comment |
1
$begingroup$
This is a better answer than Alex's but doesn't explain the significance of the number $f(x)$. Does it have a meaning independent of a cdf? Andre's answer of it being approximately $hf(a)$ is great but he doesn't indicate if there's more to $f(x)$ by itself.
$endgroup$
– Jacob
Feb 27 '13 at 17:23
1
1
$begingroup$
This is a better answer than Alex's but doesn't explain the significance of the number $f(x)$. Does it have a meaning independent of a cdf? Andre's answer of it being approximately $hf(a)$ is great but he doesn't indicate if there's more to $f(x)$ by itself.
$endgroup$
– Jacob
Feb 27 '13 at 17:23
$begingroup$
This is a better answer than Alex's but doesn't explain the significance of the number $f(x)$. Does it have a meaning independent of a cdf? Andre's answer of it being approximately $hf(a)$ is great but he doesn't indicate if there's more to $f(x)$ by itself.
$endgroup$
– Jacob
Feb 27 '13 at 17:23
add a comment |
$begingroup$
Intro statistics focuses on the PDF as the description of the population, but in fact it is the CDF (cumulative density function) that gives you a functional understanding of the population, as points on the CDF denote probabilities over a relevant range of measures. If you look at all stats from this perspective, then the PDF is just the description of probability change with respect to a change around a point along the measure at hand. The values on the PDF therefore only give you a look at the spread. For example, given two normal distributions $N(mu_1, sigma_1^2)$ and $N(mu_2, sigma_2^2)$, if you choose any value of $x$ to get point $p_n=mu_n+xcdotsigma_n$ for the respective distributions and get $X_1[p_1 ] > X_2[p_2 ]$, then this just means $sigma_1 < sigma_2$. Similar relationships exist for other distributions.
answered Jun 28 '15 at 14:44
TopherTopher
347317
$endgroup$
$begingroup$
Very interesting answer!
$endgroup$
– information_interchange
Jun 7 '18 at 18:38
add a comment |
$begingroup$
Intro statistics focuses on the PDF as the description of the population, but in fact it is the CDF (cumulative density function) that gives you a functional understanding of the population, as points on the CDF denote probabilities over a relevant range of measures. If you look at all stats from this perspective, then the PDF is just the description of probability change with respect to a change around a point along the measure at hand. The values on the PDF therefore only give you a look at the spread. For example, given two normal distributions $N(mu_1, sigma_1^2)$ and $N(mu_2, sigma_2^2)$, if you choose any value of $x$ to get point $p_n=mu_n+xcdotsigma_n$ for the respective distributions and get $X_1[p_1 ] > X_2[p_2 ]$, then this just means $sigma_1 < sigma_2$. Similar relationships exist for other distributions.
answered Jun 28 '15 at 14:44
TopherTopher
347317
$endgroup$
$begingroup$
Very interesting answer!
$endgroup$
– information_interchange
Jun 7 '18 at 18:38
add a comment |
$begingroup$
Intro statistics focuses on the PDF as the description of the population, but in fact it is the CDF (cumulative density function) that gives you a functional understanding of the population, as points on the CDF denote probabilities over a relevant range of measures. If you look at all stats from this perspective, then the PDF is just the description of probability change with respect to a change around a point along the measure at hand. The values on the PDF therefore only give you a look at the spread. For example, given two normal distributions $N(mu_1, sigma_1^2)$ and $N(mu_2, sigma_2^2)$, if you choose any value of $x$ to get point $p_n=mu_n+xcdotsigma_n$ for the respective distributions and get $X_1[p_1 ] > X_2[p_2 ]$, then this just means $sigma_1 < sigma_2$. Similar relationships exist for other distributions.
answered Jun 28 '15 at 14:44
TopherTopher
347317
$endgroup$
Intro statistics focuses on the PDF as the description of the population, but in fact it is the CDF (cumulative density function) that gives you a functional understanding of the population, as points on the CDF denote probabilities over a relevant range of measures. If you look at all stats from this perspective, then the PDF is just the description of probability change with respect to a change around a point along the measure at hand. The values on the PDF therefore only give you a look at the spread. For example, given two normal distributions $N(mu_1, sigma_1^2)$ and $N(mu_2, sigma_2^2)$, if you choose any value of $x$ to get point $p_n=mu_n+xcdotsigma_n$ for the respective distributions and get $X_1[p_1 ] > X_2[p_2 ]$, then this just means $sigma_1 < sigma_2$. Similar relationships exist for other distributions.
answered Jun 28 '15 at 14:44
TopherTopher
347317
answered Jun 28 '15 at 14:44
TopherTopher
347317
answered Jun 28 '15 at 14:44
TopherTopher
347317
answered Jun 28 '15 at 14:44
TopherTopher
347317
347317
$begingroup$
Very interesting answer!
$endgroup$
– information_interchange
Jun 7 '18 at 18:38
add a comment |
$begingroup$
Very interesting answer!
$endgroup$
– information_interchange
Jun 7 '18 at 18:38
$begingroup$
Very interesting answer!
$endgroup$
– information_interchange
Jun 7 '18 at 18:38
$begingroup$
Very interesting answer!
$endgroup$
– information_interchange
Jun 7 '18 at 18:38
add a comment |
$begingroup$
I am not sure if Jester is still interested, as it's been 5 years, but I think I found a less confusing anwer than in Wikipedia.
In contrast to discrete random variables, if X is continuous, f(X) is a function whose value at any given sample is not the probability but rather it indicates the likelihood that X will be in that sample/interval. For example if the value of the PDF around a point (can be generalized for a sample) x is large, that means the random variable X is more likely to take values close to x. If, on the other hand, f(x)=0 in some interval, then X won't be in that interval
Of course a more practical way of thinking it is that the probability of X being in an interval is given by the integral of the PDF.
You might want to look at the link below for more details: http://mathinsight.org/probability_density_function_idea
answered Oct 24 '17 at 23:25
ALEX.VAMVASALEX.VAMVAS
211
$endgroup$
add a comment |
$begingroup$
I am not sure if Jester is still interested, as it's been 5 years, but I think I found a less confusing anwer than in Wikipedia.
In contrast to discrete random variables, if X is continuous, f(X) is a function whose value at any given sample is not the probability but rather it indicates the likelihood that X will be in that sample/interval. For example if the value of the PDF around a point (can be generalized for a sample) x is large, that means the random variable X is more likely to take values close to x. If, on the other hand, f(x)=0 in some interval, then X won't be in that interval
Of course a more practical way of thinking it is that the probability of X being in an interval is given by the integral of the PDF.
You might want to look at the link below for more details: http://mathinsight.org/probability_density_function_idea
answered Oct 24 '17 at 23:25
ALEX.VAMVASALEX.VAMVAS
211
$endgroup$
add a comment |
$begingroup$
I am not sure if Jester is still interested, as it's been 5 years, but I think I found a less confusing anwer than in Wikipedia.
In contrast to discrete random variables, if X is continuous, f(X) is a function whose value at any given sample is not the probability but rather it indicates the likelihood that X will be in that sample/interval. For example if the value of the PDF around a point (can be generalized for a sample) x is large, that means the random variable X is more likely to take values close to x. If, on the other hand, f(x)=0 in some interval, then X won't be in that interval
Of course a more practical way of thinking it is that the probability of X being in an interval is given by the integral of the PDF.
You might want to look at the link below for more details: http://mathinsight.org/probability_density_function_idea
answered Oct 24 '17 at 23:25
ALEX.VAMVASALEX.VAMVAS
211
$endgroup$
I am not sure if Jester is still interested, as it's been 5 years, but I think I found a less confusing anwer than in Wikipedia.
In contrast to discrete random variables, if X is continuous, f(X) is a function whose value at any given sample is not the probability but rather it indicates the likelihood that X will be in that sample/interval. For example if the value of the PDF around a point (can be generalized for a sample) x is large, that means the random variable X is more likely to take values close to x. If, on the other hand, f(x)=0 in some interval, then X won't be in that interval
Of course a more practical way of thinking it is that the probability of X being in an interval is given by the integral of the PDF.
You might want to look at the link below for more details: http://mathinsight.org/probability_density_function_idea
answered Oct 24 '17 at 23:25
ALEX.VAMVASALEX.VAMVAS
211
answered Oct 24 '17 at 23:25
ALEX.VAMVASALEX.VAMVAS
211
answered Oct 24 '17 at 23:25
ALEX.VAMVASALEX.VAMVAS
211
answered Oct 24 '17 at 23:25
ALEX.VAMVASALEX.VAMVAS
211
211
add a comment |
add a comment |
$begingroup$
The ratio of the pdf $f(x)$ at two points, $r_x = f(x_0)/f(x_1)$, is not a measure of relative probability (or "relative likelihood") for the two outcomes for the random variable $X$. The ratio depends on the metric. That is, with a variable transformation, $z=z(x)$, with the pdf for $Z$ given by $h(z)$, the ratio $r_z=h(z_0)/h(z_1)neq r_x$, in general, even though the two ratios refer to the same two outcomes. For monotonic transformations, $f(x),dx = h(z),dz$.
Numerical values of the pdf have no value on their own. The metric, $dx$, is required for probability interpretations (ie. $f(x),dx$). Wikipedia got this wrong, so I have corrected it.
answered Nov 27 '16 at 0:47
DeanDean
1,00537
$endgroup$
add a comment |
$begingroup$
The ratio of the pdf $f(x)$ at two points, $r_x = f(x_0)/f(x_1)$, is not a measure of relative probability (or "relative likelihood") for the two outcomes for the random variable $X$. The ratio depends on the metric. That is, with a variable transformation, $z=z(x)$, with the pdf for $Z$ given by $h(z)$, the ratio $r_z=h(z_0)/h(z_1)neq r_x$, in general, even though the two ratios refer to the same two outcomes. For monotonic transformations, $f(x),dx = h(z),dz$.
Numerical values of the pdf have no value on their own. The metric, $dx$, is required for probability interpretations (ie. $f(x),dx$). Wikipedia got this wrong, so I have corrected it.
answered Nov 27 '16 at 0:47
DeanDean
1,00537
$endgroup$
add a comment |
$begingroup$
The ratio of the pdf $f(x)$ at two points, $r_x = f(x_0)/f(x_1)$, is not a measure of relative probability (or "relative likelihood") for the two outcomes for the random variable $X$. The ratio depends on the metric. That is, with a variable transformation, $z=z(x)$, with the pdf for $Z$ given by $h(z)$, the ratio $r_z=h(z_0)/h(z_1)neq r_x$, in general, even though the two ratios refer to the same two outcomes. For monotonic transformations, $f(x),dx = h(z),dz$.
Numerical values of the pdf have no value on their own. The metric, $dx$, is required for probability interpretations (ie. $f(x),dx$). Wikipedia got this wrong, so I have corrected it.
answered Nov 27 '16 at 0:47
DeanDean
1,00537
$endgroup$
The ratio of the pdf $f(x)$ at two points, $r_x = f(x_0)/f(x_1)$, is not a measure of relative probability (or "relative likelihood") for the two outcomes for the random variable $X$. The ratio depends on the metric. That is, with a variable transformation, $z=z(x)$, with the pdf for $Z$ given by $h(z)$, the ratio $r_z=h(z_0)/h(z_1)neq r_x$, in general, even though the two ratios refer to the same two outcomes. For monotonic transformations, $f(x),dx = h(z),dz$.
Numerical values of the pdf have no value on their own. The metric, $dx$, is required for probability interpretations (ie. $f(x),dx$). Wikipedia got this wrong, so I have corrected it.
answered Nov 27 '16 at 0:47
DeanDean
1,00537
answered Nov 27 '16 at 0:47
DeanDean
1,00537
answered Nov 27 '16 at 0:47
DeanDean
1,00537
answered Nov 27 '16 at 0:47
DeanDean
1,00537
1,00537
add a comment |
add a comment |
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Let $f$ be the density function of $X$. Assume $f$ is continuous. Then if $h$ is small, the probability that $X$ lies in the interval $[a,a+h]$ is approximately $hf(a)$. By approximately I mean that the probability, divided by $h$, approaches $f(a)$ as $h$ approaches $0$. So the ratio $f(a)/f(b)$ measures, approximately, the ratio of the probability that $X$ is in $[a,a+h]$ to the probability $X$ is in $[b,b+h]$.
$endgroup$
– André Nicolas
Oct 10 '12 at 19:28