Why having the loss matrix to be 1 - identity function yields a loss of $1-p(C_l|x)$ ? (Bishop exercise 1.24)
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In the solution for exercise 1.24 in Bishop's PRML solutions book I see the following statement:
For a loss matrix $L_{kj}=1-I_{kj}$ we have
$sum_kL_{kl}P(C_k|x)=1-p(C_l|x)$
Now I understand where the $1$ is coming from:
$sum_kL_{kl}P(C_k|x)=underbrace{sum_kp(C_k|x)} - sum_kp(C_k|x)$
The item in the underbrace equals $1$ because it's just the sum of the probabilities. But how is the second item equal to $p(C_l|x)$ ?
probability matrices machine-learning
asked Dec 1 at 13:10
Alaa M.
15717
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In the solution for exercise 1.24 in Bishop's PRML solutions book I see the following statement:
For a loss matrix $L_{kj}=1-I_{kj}$ we have
$sum_kL_{kl}P(C_k|x)=1-p(C_l|x)$
Now I understand where the $1$ is coming from:
$sum_kL_{kl}P(C_k|x)=underbrace{sum_kp(C_k|x)} - sum_kp(C_k|x)$
The item in the underbrace equals $1$ because it's just the sum of the probabilities. But how is the second item equal to $p(C_l|x)$ ?
probability matrices machine-learning
asked Dec 1 at 13:10
Alaa M.
15717
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the solution for exercise 1.24 in Bishop's PRML solutions book I see the following statement:
For a loss matrix $L_{kj}=1-I_{kj}$ we have
$sum_kL_{kl}P(C_k|x)=1-p(C_l|x)$
Now I understand where the $1$ is coming from:
$sum_kL_{kl}P(C_k|x)=underbrace{sum_kp(C_k|x)} - sum_kp(C_k|x)$
The item in the underbrace equals $1$ because it's just the sum of the probabilities. But how is the second item equal to $p(C_l|x)$ ?
probability matrices machine-learning
asked Dec 1 at 13:10
Alaa M.
15717
In the solution for exercise 1.24 in Bishop's PRML solutions book I see the following statement:
For a loss matrix $L_{kj}=1-I_{kj}$ we have
$sum_kL_{kl}P(C_k|x)=1-p(C_l|x)$
Now I understand where the $1$ is coming from:
$sum_kL_{kl}P(C_k|x)=underbrace{sum_kp(C_k|x)} - sum_kp(C_k|x)$
The item in the underbrace equals $1$ because it's just the sum of the probabilities. But how is the second item equal to $p(C_l|x)$ ?
probability matrices machine-learning
probability matrices machine-learning
asked Dec 1 at 13:10
Alaa M.
15717
asked Dec 1 at 13:10
Alaa M.
15717
asked Dec 1 at 13:10
Alaa M.
15717
asked Dec 1 at 13:10
Alaa M.
15717
asked Dec 1 at 13:10
Alaa M.
15717
15717
add a comment |
add a comment |
1 Answer
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The second term is actually $sum_kcolor{red}{I_{kl}}p(C_k|x)$ which equals $p(C_l|x)$ as $I_{kl}=1$ only if $k=l$, else $I_{kl}$ is zero.
answered Dec 1 at 13:31
Alijah Ahmed
9,86421217
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The second term is actually $sum_kcolor{red}{I_{kl}}p(C_k|x)$ which equals $p(C_l|x)$ as $I_{kl}=1$ only if $k=l$, else $I_{kl}$ is zero.
answered Dec 1 at 13:31
Alijah Ahmed
9,86421217
add a comment |
up vote
1
down vote
accepted
The second term is actually $sum_kcolor{red}{I_{kl}}p(C_k|x)$ which equals $p(C_l|x)$ as $I_{kl}=1$ only if $k=l$, else $I_{kl}$ is zero.
answered Dec 1 at 13:31
Alijah Ahmed
9,86421217
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The second term is actually $sum_kcolor{red}{I_{kl}}p(C_k|x)$ which equals $p(C_l|x)$ as $I_{kl}=1$ only if $k=l$, else $I_{kl}$ is zero.
answered Dec 1 at 13:31
Alijah Ahmed
9,86421217
The second term is actually $sum_kcolor{red}{I_{kl}}p(C_k|x)$ which equals $p(C_l|x)$ as $I_{kl}=1$ only if $k=l$, else $I_{kl}$ is zero.
answered Dec 1 at 13:31
Alijah Ahmed
9,86421217
answered Dec 1 at 13:31
Alijah Ahmed
9,86421217
answered Dec 1 at 13:31
Alijah Ahmed
9,86421217
answered Dec 1 at 13:31
Alijah Ahmed
9,86421217
9,86421217
add a comment |
add a comment |
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