# Combination of probabilities for probability densities

If I have an experiment with discrete probabilities like drawing black and white balls, the probability of drawing both black and white is just a multiplication of probabilities of drawing black and white. Similarly, if I want to work with "OR" type of situations, I add the probabilities.

But how does this work with continuous random variables? Say, I have the following experiment: I have a large bucket of some material and I am trying to determine the melt temperature. So I take small samples and put them on a burner and note the temperature when they melt. Let's assume that after many of such measurements, I get a gaussian with some mean value and variance.

If I then would like to say something about another bucket of the same material, I might ask, what is the probability of a sample melting at temperature between $(T,T+{\rm d}T)$. Then the probability should be $\Pr(\text{melt at }T)\approx f(T)\,{\rm d} T$ or more exactly $\int_T^{T+{\rm d}T}f(x)\,{\rm d} x$ where $f(T)$ is the gaussian.

But what if I want to ask about two samples from taken from the bucket both melting at $\{T,T+{\rm d}T\}$ or one of the samples from the bucket melting between $(T_1,T_1+{\rm d}T)$ and the other from the bucket melting between $(T_2,T_2+{\rm d}T)$? Is it just $f^2(T)\,{\rm d}x^2$ and $f(T_1)f(T_2)\,{\rm d}x^2$ respectively? Or more exactly

$$\int_T^{T+{\rm d}T}f(x)\,{\rm d} x\int_T^{T+{\rm d}T}f(x)\,{\rm d} x$$

or

$$\int_{T_1}^{T_1+{\rm d}T} f(x){\rm d} x \int_{T_2}^{T_2+{\rm d}T}f(x)\,{\rm d} x$$

But this leads to having new probability distribution functions $f(x)f(y)$ which from what I have read are not necessarily distribution functions.

And if I want to ask about the probability of the melt temperature being between $(T_1, T_1+{\rm d}T)$ or between $(T_2,T_2+{\rm d}T)$, would it be $f(T_1)\,{\rm d}T + f(T_2)\,{\rm d}T$ or more precisely

$$\int_{T_1}^{T_1+{\rm d}T} f(x)\,{\rm d} x + \int_{T_2}^{T_2+{\rm d}T} f(x)\,{\rm d} x\text{?}$$

• You should ask this question at stats.stackexchange.com and google "Analysis of variance". From multiple samples you can estimate the necessary parameters and then estimate such probabilities with associated measures of precision. – JimB Dec 26 '17 at 16:11
• @JimB No. The OP's trouble is definitely caused by a misunderstanding of the basics of probability theory, their mention of samples being anecdotal. (And I have no idea why you think that analysis of variance would be relevant.) – Did Dec 28 '17 at 16:03
• @Did Would you be so kind and suggest what my basic misunderstanding is and what do you find anecdotal about samples? – leosenko Dec 28 '17 at 16:26
• @Did ANOVA: taking samples from a batch of material and measuring temperatures. Maybe conditions change from batch to batch either in a random way or there is some associated factor. Then with estimates of the mean and variance components one can estimate probabilities of a future batch having a melting temperature between two particular values. – JimB Dec 28 '17 at 16:56
• @Did I guess what I'm trying to say is that the parameters of some model need to be estimated from samples and then functions of the estimated parameters can be examined (and providing along the way some estimate of precision). It's not about starting out with known parameters. – JimB Dec 28 '17 at 17:09

I think that there is an issue about terminology (which is my feeble defense about my earlier interpretation of the problem).

If the temperature of a single sample can be considered to have a Gaussian distribution with mean $\mu$ and variance $\sigma^2$, then the probability of the observed temperature being between $T$ and $T+dT$ is given by

$$p = \Phi\left({{T+dT-\mu}\over{\sigma}}\right) - \Phi\left({{T-\mu}\over{\sigma}}\right)$$

The probability of two independent samples both being between $T$ and $T+dT$ is the same temperature range is the square of the above ($p^2$). If the samples are not independent, then you'd need to define how they are dependent. Also, this is not the same answer to the question "What is the probability that both samples will be within $dT$ of each other?"

Still assuming independence of samples and that you know what the parameters are for the Gaussian distribution, the probability that sample 1 is between $T_1$ and $T_1+dT$ and that sample 2 is between $T_2$ and $T_2+dT$ is just the product the two probabilities.

If now you want to know the probability of a single sample resulting in a temperature between $T_1$ and $T_1+dT$ or $T_2$ and $T_2+dT$, then the probability of that event occurring is just the sum of the two probabilities if there is no overlap in the two intervals.

Now suppose there is overlap with $T_1 \le T_2 \le T_1+dT$. Then the probability is

$$\Phi\left({{T_2+dT-\mu}\over{\sigma}}\right) - \Phi\left({{T_1-\mu}\over{\sigma}}\right)$$

But all of the above assumes that you know $\mu$ and $\sigma^2$. If you just have estimates of $\mu$ and $\sigma^2$ from previous samples, then you can certainly estimate the above probabilities but you'd also want to estimate some associated measure of precision. That would require a bit more explanation.

• Thank you very much. You mention the dependence of the samples making me think what such a dependence would be? I believe that the independence is in the sense that the melt temperature of one is not influenced by the other, correct? Another thing that I am not sure: why do you need no overlap between the two temperature ranges in your 5th paragraph? I am guessing that this is because of $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ where $A$ and $B$ are the events of melting at some $T$.But the fact that one melts at some range should have no influence at the range for the second sample,no? – leosenko Dec 31 '17 at 14:00
• I think you are confusing the events $A$ and $B$ with the two independent samples. It's all about being explicit about the definition of events. Let $A_i$ be the event that sample $i$ has a melting temperature between $T_A$ and $T_A+dT$ and $B_i$ be the event that sample $i$ has a melting temperature between $T_B$ and $T_B+dT$. Then because we have independence of samples, $A_1$ is independent of both $A_2$ and $B_2$ but $A_1$ is not independent of $B_1$. – JimB Dec 31 '17 at 16:54
• If $T_A \le T_B \le T_A+dT$ or $T_B \le T_A \le T_B+dT$, then $A_i$ and $B_i$ are not disjoint events and the probability of the union of those events is not just the sum of their individual probabilities but rather $P(A_i)+P(B_i)-P(A_i\cap B_i)$. – JimB Dec 31 '17 at 16:55
• Thank you for all the explanation. – leosenko Jan 1 '18 at 7:56