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Tim
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Agree with you, this quote is confusing. Probability is bounded between zero and one, probability density is non-negative, so "positive probability" taken literally means non-zero. My guess would be that by "positive probability" they mean something like "high probability" and state the tautology "for $x$ such that probability density $p(x)$ is high, the probability of observing value in close proximity to $x$ is high". Otherwise they would be suggesting that probability of observing some $x$ value does not correspond to $p(x)$, what is nonsense.

Another interpretation might be that authors are trying to define probability density in here. For continuous variable $\Pr(X=x) = 0$ for any $x$, so we use probability densities instead. Probability densities $p(x)$ are "probabilities per foot", so $\int_a^b p(x) \, dx = \Pr(a < x < b)$ tells us about probability of observing $x$ within the $(a,b)$ range. You could say something similar like the authors: "if probability density $p(x)$ is large, there's high probability, the feature vector $x$ will fall near to $x$".

Agree with you, this quote is confusing. Probability is bounded between zero and one, probability density is non-negative, so "positive probability" taken literally means non-zero. My guess would be that by "positive probability" they mean something like "high probability" and state the tautology "for $x$ such that $p(x)$ is high, the probability of observing $x$ is high". Otherwise they would be suggesting that probability of observing some $x$ value does not correspond to $p(x)$, what is nonsense.

Agree with you, this quote is confusing. Probability is bounded between zero and one, probability density is non-negative, so "positive probability" taken literally means non-zero. My guess would be that by "positive probability" they mean something like "high probability" and state the tautology "for $x$ such that probability density $p(x)$ is high, the probability of observing value in close proximity to $x$ is high". Otherwise they would be suggesting that probability of observing some $x$ value does not correspond to $p(x)$, what is nonsense.

Another interpretation might be that authors are trying to define probability density in here. For continuous variable $\Pr(X=x) = 0$ for any $x$, so we use probability densities instead. Probability densities $p(x)$ are "probabilities per foot", so $\int_a^b p(x) \, dx = \Pr(a < x < b)$ tells us about probability of observing $x$ within the $(a,b)$ range. You could say something similar like the authors: "if probability density $p(x)$ is large, there's high probability, the feature vector $x$ will fall near to $x$".

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Tim
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Agree with you, this quote is confusing. Probability is bounded between zero and one, probability density is non-negative, so "positive probability" taken literally means non-zero. My guess would be that by "positive probability" they mean something like "high probability" and state the tautology "for $x$ such that $p(x)$ is high, the probability of observing $x$ is high". Otherwise they would be suggesting that probability of observing some $x$ value can differ fromdoes not correspond to $p(x)$, what is nonsense.

Agree with you, this quote is confusing. Probability is bounded between zero and one, probability density is non-negative, so "positive probability" taken literally means non-zero. My guess would be that by "positive probability" they mean something like "high probability" and state the tautology "for $x$ such that $p(x)$ is high, the probability of observing $x$ is high". Otherwise they would be suggesting that probability of observing some $x$ value can differ from $p(x)$, what is nonsense.

Agree with you, this quote is confusing. Probability is bounded between zero and one, probability density is non-negative, so "positive probability" taken literally means non-zero. My guess would be that by "positive probability" they mean something like "high probability" and state the tautology "for $x$ such that $p(x)$ is high, the probability of observing $x$ is high". Otherwise they would be suggesting that probability of observing some $x$ value does not correspond to $p(x)$, what is nonsense.

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Tim
  • 141.2k
  • 26
  • 270
  • 512

Agree with you, this quote is confusing. Probability is bounded between zero and one, probability density is non-negative, so "positive probability" taken literally means non-zero. My guess would be that by "positive probability" they mean something like "high probability" and state the tautology "for $x$ such that $p(x)$ is high, the probability of observing $x$ is high". Otherwise they would be suggesting that probability of observing some $x$ value can differ from $p(x)$, what is nonsense.