# Connection between PDFs/PMFs and Bayes Theorem

UPDATE Original question was confused and poorly worded. I thought about it more and don't think I have a question any longer. After thinking a bit more I came up with:

For a distribution, such as the normal for example, the probability that the variable falls in a certain range is:

P(a<=X<=b|X ~ N(mu, sigma^2))


Using Bayes' Theorem I can infer:

 P(a<=X<=b|X ~ N(mu, sigma^2)) = P(a<=X<=b, X ~ N(mu, sigma^2)) / P(X ~ N(mu, sigma^2))


Interpreting this statement I could say:

P(X ~ N(mu, sigma^2)) is the prior probability that the random variable X ~ N(mu, sigma^2) (based on whatever information we have), and P(a<=X<=b, X ~ N(mu, sigma^2)) is the prior probability that X ~ N(mu, sigma^2) and within the interval [a,b]. Then, if it is revealed to us that X is in fact distributed N(mu, sigma^2), or if we assume it to be so, the probability that it falls within the interval [a,b] is the ratio of the prior probabilities, as stated above.

The above statement doesn't depend at all on what the particular pdf is that you're dealing with. I think I was hoping to see a connection between the two, but they seem to be unrelated ideas.

• >Does this make any sense? No, you need to understand some basic ideas about conditional probability. Commented Jul 25, 2013 at 0:42
• You need to distinguish between two different uses of the word "given". In a problem such as $$\text{Given}~P(A) = 0.3, ~\text{find} ~ P(A^c)$$ you are not being asked for a conditional probability. Here, given is giving you information, not saying that something happened. In a problem such as $$\text{Given that}~k~ \text{Heads occurred on}~n~\text{tosses of a coin, what is the probability that the first toss resulted in a Head?}$$ you are being asked for a conditional probability. The key is to look for verbs such as occurred that tell you that something happened. Commented Jul 25, 2013 at 21:50
• @Dilip Sarwate, thank you for the clarification. So, to confirm, is it correct to say that in your first example that you would not condition because the information we have: P(A) = 0.3, is not an event? On the other hand, in the second case the information provided: "k Heads occurred on n tosses of a coin," is describing an event, so we can condition on it. Applying this to my original problem, then, the reasoning would be that X ~ N(mu, sigma^2) is a statement, not an event, so conditioning on it is an incorrect application of conditional probability. Right? Commented Jul 25, 2013 at 23:19
• Yes, you are getting to understand the distinction. Commented Jul 26, 2013 at 2:36

Suppose $f_{X}(x)$ is the pdf of a random variable $X$. Then $$P(a \leq X \leq b) = \int_{a}^{b} f_{X}(x) \ dx$$
The probability of $X =c$ is $0$ where $c \in [a,b]$. The value of $f_{X}(x)$ is not a probability.