I have a set where its values follow a normal distribution, but I only have the sum of all of them, and the sum of its squares.

Having: $\sum_{i=0}^n X_i$ and $\sum_{i=0}^n X_i^2$.

I have the following "table":

((+ (fever 245 5975))
 (- (fever 176 4540)))

There are two types of classes: Positive (+) and Negative(-). For both, I have the sum (245 and 176), and the sum of their squares (5975 and 4540).

How can I calculate P(Positive class given 45)?

P(Positive class given 45) or $P(Positive|45)$ means that given a fever of 45, it will be positive case and the patient will have measles. This is an invented example.

I think I have to use the mean and the variance, but I don't know how.


closed as off-topic by callculus, dantopa, clathratus, Santana Afton, Lee David Chung Lin Mar 27 at 2:41

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  • $\begingroup$ What does "$P(Positive|25)$" mean? $\endgroup$ – robjohn Mar 25 at 16:21
  • $\begingroup$ I have updated my question. $\endgroup$ – VansFannel Mar 25 at 16:26
  • $\begingroup$ Sorry, but your table is unreadable. Please fix it. Additional it seems that you´ve omitted some context. $\endgroup$ – callculus Mar 25 at 16:38
  • $\begingroup$ @callculus I don't think so. I'm only asking if it is possible to calculate conditional probability with the mean and the variance. $\endgroup$ – VansFannel Mar 25 at 16:44
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    $\begingroup$ This question is really unclear to me. What do the $X_i$ represent? What does $n$ represent? I'm guessing that the positive class is actually positive for some condition (say, measles), and the negative class is actually negative for that condition; is that right? Why do you think that the $X_i$ are normally distributed? (You probably shouldn't use "fever" as an example, since a fever of $45$ means DEAD whether you're talking F or C.) $\endgroup$ – Brian Tung Mar 25 at 17:14

I'm going to hazard a guess as to what you mean, in ordinary terms. I'm going to guess that you have two normal distributions, with estimated means $\mu_c$ (class $c = 1, 2$) and standard deviations $\sigma_c$. These can be derived from the sum and sum of squares, provided that you know $n$ (for each class, if they're different).

The estimated mean is

$$ \mu = \frac1n \sum_{i=1}^n X_i $$

and the estimated standard deviation is

$$ \sigma = \sqrt{\frac{\left(\sum_{i=1}^n X_i^2\right) - n\mu^2}{n-1}} $$

We can then use the PDF of the normal distribution at a given value $x$ to compute the relative likelihoods of being positive or negative for the condition:

$$ f_c(x) = \frac{1}{\sqrt{2\pi\sigma_c^2}}\exp\left[-\frac{(x-\mu_c)^2}{2\sigma_c^2}\right] $$

Then I think the probability you want is

$$ P(\text{class $c$} \mid x) = \frac{p_cf_c(x)}{p_1f_1(x)+p_2f_2(x)} \qquad c = 1, 2 $$

where $p_c$ is the a priori probability of being in class $c$.

  • 1
    $\begingroup$ Probably the right guess. :) A possible extra twist is I wouldn't be surprised if the no. of samples in each class is different, i.e. $n_1, n_2$... Then you can either use a likelihood approach, or even consider the ratio $n_1 : n_2$ to be a prior and use Bayesian. BTW your final formula ... wouldn't it evaluate to $>1$ in some cases? $\endgroup$ – antkam Mar 25 at 17:35
  • $\begingroup$ @antkam: Oops, yeah, I forgot the $p_c$! Thanks for the catch. Incidentally, I didn't think of using $n_1 : n_2$ as a prior, since I think it's not likely to be appropriate. But perhaps... $\endgroup$ – Brian Tung Mar 25 at 17:38
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    $\begingroup$ Who said anything about appropriate? This sounds like a game of "lets guess what the teacher / examiner had in mind" ;) $\endgroup$ – antkam Mar 25 at 17:56

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