Mixing
probability of belonging to a class if distribution of classes are known
Clash Royale CLAN TAG#URR8PPP
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Consider a population of samples that depend on a variable $X$ and are distributed between one of 2 classes (e.g. Green and Red like this).
The probability density functions $p_G(x)$ and $p_R(x)$ for each of the 2 classes are known.
Given a value for $X=x$, what is the probability $P(G|X=x)$ or $P(R|X=x)$ that a sample belongs to one or the other class ?
From here, I suspect that: $$P(G|X=x) = fracp_G(x)p_G(x)+p_R(x)$$
But how can it be proved formally ?
probability density-function
asked Jul 25 at 16:46
guik
11
add a comment |Â
up vote
0
down vote
favorite
Consider a population of samples that depend on a variable $X$ and are distributed between one of 2 classes (e.g. Green and Red like this).
The probability density functions $p_G(x)$ and $p_R(x)$ for each of the 2 classes are known.
Given a value for $X=x$, what is the probability $P(G|X=x)$ or $P(R|X=x)$ that a sample belongs to one or the other class ?
From here, I suspect that: $$P(G|X=x) = fracp_G(x)p_G(x)+p_R(x)$$
But how can it be proved formally ?
probability density-function
asked Jul 25 at 16:46
guik
11
2
If initially before observing $x$ you had Red and Green being equally likely, then your result is an application of Bayes' theorem. Incidentally it is usual to write $X=x$ where $X$ is the random variable and $x$ the observed value, making your formula $mathbb P(Gmid X=x) = dfracp_G(x)p_G(x)+p_R(x)$
– Henry
Jul 26 at 0:22
Thank you, I have updated the notation, and your comment about Baye's theorem is indeed the answer.
– guik
Jul 26 at 9:14
As another minor comment, Thomas Bayes had Ss at the ends of his names
– Henry
Jul 26 at 9:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider a population of samples that depend on a variable $X$ and are distributed between one of 2 classes (e.g. Green and Red like this).
The probability density functions $p_G(x)$ and $p_R(x)$ for each of the 2 classes are known.
Given a value for $X=x$, what is the probability $P(G|X=x)$ or $P(R|X=x)$ that a sample belongs to one or the other class ?
From here, I suspect that: $$P(G|X=x) = fracp_G(x)p_G(x)+p_R(x)$$
But how can it be proved formally ?
probability density-function
asked Jul 25 at 16:46
guik
11
Consider a population of samples that depend on a variable $X$ and are distributed between one of 2 classes (e.g. Green and Red like this).
The probability density functions $p_G(x)$ and $p_R(x)$ for each of the 2 classes are known.
Given a value for $X=x$, what is the probability $P(G|X=x)$ or $P(R|X=x)$ that a sample belongs to one or the other class ?
From here, I suspect that: $$P(G|X=x) = fracp_G(x)p_G(x)+p_R(x)$$
But how can it be proved formally ?
probability density-function
asked Jul 25 at 16:46
guik
11
edited Jul 26 at 9:10
asked Jul 25 at 16:46
guik
11
asked Jul 25 at 16:46
guik
11
asked Jul 25 at 16:46
guik
11
11
2
If initially before observing $x$ you had Red and Green being equally likely, then your result is an application of Bayes' theorem. Incidentally it is usual to write $X=x$ where $X$ is the random variable and $x$ the observed value, making your formula $mathbb P(Gmid X=x) = dfracp_G(x)p_G(x)+p_R(x)$
– Henry
Jul 26 at 0:22
Thank you, I have updated the notation, and your comment about Baye's theorem is indeed the answer.
– guik
Jul 26 at 9:14
As another minor comment, Thomas Bayes had Ss at the ends of his names
– Henry
Jul 26 at 9:36
add a comment |Â
2
If initially before observing $x$ you had Red and Green being equally likely, then your result is an application of Bayes' theorem. Incidentally it is usual to write $X=x$ where $X$ is the random variable and $x$ the observed value, making your formula $mathbb P(Gmid X=x) = dfracp_G(x)p_G(x)+p_R(x)$
– Henry
Jul 26 at 0:22
Thank you, I have updated the notation, and your comment about Baye's theorem is indeed the answer.
– guik
Jul 26 at 9:14
As another minor comment, Thomas Bayes had Ss at the ends of his names
– Henry
Jul 26 at 9:36
2
2
If initially before observing $x$ you had Red and Green being equally likely, then your result is an application of Bayes' theorem. Incidentally it is usual to write $X=x$ where $X$ is the random variable and $x$ the observed value, making your formula $mathbb P(Gmid X=x) = dfracp_G(x)p_G(x)+p_R(x)$
– Henry
Jul 26 at 0:22
If initially before observing $x$ you had Red and Green being equally likely, then your result is an application of Bayes' theorem. Incidentally it is usual to write $X=x$ where $X$ is the random variable and $x$ the observed value, making your formula $mathbb P(Gmid X=x) = dfracp_G(x)p_G(x)+p_R(x)$
– Henry
Jul 26 at 0:22
Thank you, I have updated the notation, and your comment about Baye's theorem is indeed the answer.
– guik
Jul 26 at 9:14
Thank you, I have updated the notation, and your comment about Baye's theorem is indeed the answer.
– guik
Jul 26 at 9:14
As another minor comment, Thomas Bayes had Ss at the ends of his names
– Henry
Jul 26 at 9:36
As another minor comment, Thomas Bayes had Ss at the ends of his names
– Henry
Jul 26 at 9:36
add a comment |Â
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2
If initially before observing $x$ you had Red and Green being equally likely, then your result is an application of Bayes' theorem. Incidentally it is usual to write $X=x$ where $X$ is the random variable and $x$ the observed value, making your formula $mathbb P(Gmid X=x) = dfracp_G(x)p_G(x)+p_R(x)$
– Henry
Jul 26 at 0:22
Thank you, I have updated the notation, and your comment about Baye's theorem is indeed the answer.
– guik
Jul 26 at 9:14
As another minor comment, Thomas Bayes had Ss at the ends of his names
– Henry
Jul 26 at 9:36