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Marginalization of joint distribution
2019 Community Moderator Electionpredicting probability distribution for time seriesTesting fit of probability distributionAn unbiased simulator for policy simulation in reinforcement learningMultimodal distribution and GANsHow does binary cross entropy work?Calibrate the predicted class probability to make it represent a true probability?How do I combine two electromagnetic readings to predict the position of a sensor?the probability distribution of dependent variablesWavenet joint probabilityHow to elegantly caclulate probability distribution parameters for a particular random variable given some observed data?
$begingroup$
I am trying to understand how you marginalise a joint distribution.
In my case I have a fair coin, $P(C) = frac12$ and a fair dice $P(D) = frac16$. I am told I win a prize if I flip the coin and it lands on Tails and if the outcome of the dice $= 1$. I am told at least one of them is correct.
$$Q = (textCoin = Tails or Dice = 1)$$
$$W = (textCoin = Tails and Dice = 1)$$
So if I wanted to work out the probability $W =$ True $| Q =$ True I can use marginalisation to work this out given the joint distribution:
$$P(C), P(D), P(Q|C,D), P(W|C,D)$$
I am just not sure where to start any help would be really appreciated. I am pretty new to this.
Thanks in advance.
probability bayesian-networks
edited Feb 25 at 18:20
Siong Thye Goh
1,387520
asked Feb 25 at 13:22
Jackt153Jackt153
111
$endgroup$
bumped to the homepage by Community♦ 29 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
I am trying to understand how you marginalise a joint distribution.
In my case I have a fair coin, $P(C) = frac12$ and a fair dice $P(D) = frac16$. I am told I win a prize if I flip the coin and it lands on Tails and if the outcome of the dice $= 1$. I am told at least one of them is correct.
$$Q = (textCoin = Tails or Dice = 1)$$
$$W = (textCoin = Tails and Dice = 1)$$
So if I wanted to work out the probability $W =$ True $| Q =$ True I can use marginalisation to work this out given the joint distribution:
$$P(C), P(D), P(Q|C,D), P(W|C,D)$$
I am just not sure where to start any help would be really appreciated. I am pretty new to this.
Thanks in advance.
probability bayesian-networks
edited Feb 25 at 18:20
Siong Thye Goh
1,387520
asked Feb 25 at 13:22
Jackt153Jackt153
111
$endgroup$
bumped to the homepage by Community♦ 29 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
I am trying to understand how you marginalise a joint distribution.
In my case I have a fair coin, $P(C) = frac12$ and a fair dice $P(D) = frac16$. I am told I win a prize if I flip the coin and it lands on Tails and if the outcome of the dice $= 1$. I am told at least one of them is correct.
$$Q = (textCoin = Tails or Dice = 1)$$
$$W = (textCoin = Tails and Dice = 1)$$
So if I wanted to work out the probability $W =$ True $| Q =$ True I can use marginalisation to work this out given the joint distribution:
$$P(C), P(D), P(Q|C,D), P(W|C,D)$$
I am just not sure where to start any help would be really appreciated. I am pretty new to this.
Thanks in advance.
probability bayesian-networks
edited Feb 25 at 18:20
Siong Thye Goh
1,387520
asked Feb 25 at 13:22
Jackt153Jackt153
111
$endgroup$
I am trying to understand how you marginalise a joint distribution.
In my case I have a fair coin, $P(C) = frac12$ and a fair dice $P(D) = frac16$. I am told I win a prize if I flip the coin and it lands on Tails and if the outcome of the dice $= 1$. I am told at least one of them is correct.
$$Q = (textCoin = Tails or Dice = 1)$$
$$W = (textCoin = Tails and Dice = 1)$$
So if I wanted to work out the probability $W =$ True $| Q =$ True I can use marginalisation to work this out given the joint distribution:
$$P(C), P(D), P(Q|C,D), P(W|C,D)$$
I am just not sure where to start any help would be really appreciated. I am pretty new to this.
Thanks in advance.
probability bayesian-networks
probability bayesian-networks
edited Feb 25 at 18:20
Siong Thye Goh
1,387520
asked Feb 25 at 13:22
Jackt153Jackt153
111
edited Feb 25 at 18:20
Siong Thye Goh
1,387520
asked Feb 25 at 13:22
Jackt153Jackt153
111
edited Feb 25 at 18:20
Siong Thye Goh
1,387520
edited Feb 25 at 18:20
Siong Thye Goh
1,387520
edited Feb 25 at 18:20
Siong Thye Goh
1,387520
1,387520
asked Feb 25 at 13:22
Jackt153Jackt153
111
asked Feb 25 at 13:22
Jackt153Jackt153
111
asked Feb 25 at 13:22
Jackt153Jackt153
111
111
bumped to the homepage by Community♦ 29 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 29 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Guide:
To solve the problem, we have to assume that there is the outcome of the dice and the outcome of the coins are independent.
We let $C$ denotes the event that the coin lands on tail and $D$ be the event that the dice outcome is $1$.
You have been given that the coin lands on tails or the dice lands on $1$, and you are interested in finding out that the coin lands on tails and the dice lands on $1$.
You want to compute $P(W|Q)=fracP(Wcap Q)P(Q)=fracP(W)P(Q)=fracP(C)P(D)P(Q)$.
To compute $P(Q)$ where $Q= C cup D$. You can either use $$P(Q)= P(C)+P(D)-P(Ccap D)$$
or $$P(Q)=1-P(Q^c)=1-P(C^c cap D^c)$$
Given all these formulas, hopefully you can solve for $P(W|Q)$.
answered Feb 25 at 16:30
Siong Thye GohSiong Thye Goh
1,387520
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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$begingroup$
Guide:
To solve the problem, we have to assume that there is the outcome of the dice and the outcome of the coins are independent.
We let $C$ denotes the event that the coin lands on tail and $D$ be the event that the dice outcome is $1$.
You have been given that the coin lands on tails or the dice lands on $1$, and you are interested in finding out that the coin lands on tails and the dice lands on $1$.
You want to compute $P(W|Q)=fracP(Wcap Q)P(Q)=fracP(W)P(Q)=fracP(C)P(D)P(Q)$.
To compute $P(Q)$ where $Q= C cup D$. You can either use $$P(Q)= P(C)+P(D)-P(Ccap D)$$
or $$P(Q)=1-P(Q^c)=1-P(C^c cap D^c)$$
Given all these formulas, hopefully you can solve for $P(W|Q)$.
answered Feb 25 at 16:30
Siong Thye GohSiong Thye Goh
1,387520
$endgroup$
add a comment |
$begingroup$
Guide:
To solve the problem, we have to assume that there is the outcome of the dice and the outcome of the coins are independent.
We let $C$ denotes the event that the coin lands on tail and $D$ be the event that the dice outcome is $1$.
You have been given that the coin lands on tails or the dice lands on $1$, and you are interested in finding out that the coin lands on tails and the dice lands on $1$.
You want to compute $P(W|Q)=fracP(Wcap Q)P(Q)=fracP(W)P(Q)=fracP(C)P(D)P(Q)$.
To compute $P(Q)$ where $Q= C cup D$. You can either use $$P(Q)= P(C)+P(D)-P(Ccap D)$$
or $$P(Q)=1-P(Q^c)=1-P(C^c cap D^c)$$
Given all these formulas, hopefully you can solve for $P(W|Q)$.
answered Feb 25 at 16:30
Siong Thye GohSiong Thye Goh
1,387520
$endgroup$
add a comment |
$begingroup$
Guide:
To solve the problem, we have to assume that there is the outcome of the dice and the outcome of the coins are independent.
We let $C$ denotes the event that the coin lands on tail and $D$ be the event that the dice outcome is $1$.
You have been given that the coin lands on tails or the dice lands on $1$, and you are interested in finding out that the coin lands on tails and the dice lands on $1$.
You want to compute $P(W|Q)=fracP(Wcap Q)P(Q)=fracP(W)P(Q)=fracP(C)P(D)P(Q)$.
To compute $P(Q)$ where $Q= C cup D$. You can either use $$P(Q)= P(C)+P(D)-P(Ccap D)$$
or $$P(Q)=1-P(Q^c)=1-P(C^c cap D^c)$$
Given all these formulas, hopefully you can solve for $P(W|Q)$.
answered Feb 25 at 16:30
Siong Thye GohSiong Thye Goh
1,387520
$endgroup$
Guide:
To solve the problem, we have to assume that there is the outcome of the dice and the outcome of the coins are independent.
We let $C$ denotes the event that the coin lands on tail and $D$ be the event that the dice outcome is $1$.
You have been given that the coin lands on tails or the dice lands on $1$, and you are interested in finding out that the coin lands on tails and the dice lands on $1$.
You want to compute $P(W|Q)=fracP(Wcap Q)P(Q)=fracP(W)P(Q)=fracP(C)P(D)P(Q)$.
To compute $P(Q)$ where $Q= C cup D$. You can either use $$P(Q)= P(C)+P(D)-P(Ccap D)$$
or $$P(Q)=1-P(Q^c)=1-P(C^c cap D^c)$$
Given all these formulas, hopefully you can solve for $P(W|Q)$.
answered Feb 25 at 16:30
Siong Thye GohSiong Thye Goh
1,387520
answered Feb 25 at 16:30
Siong Thye GohSiong Thye Goh
1,387520
answered Feb 25 at 16:30
Siong Thye GohSiong Thye Goh
1,387520
answered Feb 25 at 16:30
Siong Thye GohSiong Thye Goh
1,387520
1,387520
add a comment |
add a comment |
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