# Proving the product rule using probability

• I
• Office_Shredder
In summary, the conversation discusses a proof of the product rule using probability theory. It shows that the probability density function of the maximum of two independent random variables can be calculated by multiplying the individual probability density functions and adding them together. This approach can be applied to any functions as long as they meet certain requirements. The conversation also mentions an intuitive argument for the product rule using a rectangle with side lengths of the individual functions.

#### Office_Shredder

Staff Emeritus
Gold Member
I thought this was kind of a cool proof of the product rule.

Let ##F(x)## and ##G(x)## be cumulative distribution functions for independent random variables ##A## and ##B## respectively with probability density functions ##f(x)=F'(x)##, ##g(x)=G'(x)##. Consider the random variable ##C=\max(A,B)##. Let ##H(x)## be the cumulative distribution function of ##C##, with pdf ##h(x)=H'(x)##. Then ##h(x) = f(x)G(x) + F(x)g(x)##. If ##C=x##, it's because either ##A=x## and ##B\leq x## or ##B=x## and ##A\leq x##. But since ##A## and ##B## are independent, to both be smaller than ##x##, the probabilities just multiply. So the cdf is simply ##F(x)G(x)##. Therefore ##\frac{d}{dx} \left(F(x)G(x)\right) = F'(x)G(x) + F(x)G'(x)##.

That's pretty much it! I thought it was kind of neat and wanted to share it.

vela, FactChecker, etotheipi and 2 others
Having done the proof via probability, does that limit its scope to only probabilistic functions?

member 587159 and FactChecker
The direct proof is simple enough ##H(x)=P(C\le x)=P(A\le x, B\le x)=P(A\le x)P(B\le x)=F(x)G(x)## The last step uses independence of ##A## and ##B##.

I wondered how you deduced ##h(x)## without going via. the CDF first? Usually to derive the PDF of the ##\text{max}## function I would have thought you say, assuming independence, ##H(x) = P(C \leq x) = P(A \leq x) P(B \leq x) = F(x)G(x)## and then make use of the product rule to find ##h(x)##.

But for this proof we need to do it backward, i.e. already know what ##h(x)## is. So is there another way of obtaining ##h(x)##? Thanks... sorry if I missed the point !

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jedishrfu said:
Having done the proof via probability, does that limit its scope to only probabilistic functions?

Technically here F and G are increasing functions between 0 and 1. But the derivative scales with multiplication by a constant, and is invariant under constant shifts, and also is only a property of the function locally. So I think it's easy to prove given any F and G, you can shift them and maybe multiply them by -1 and then redefine them everywhere except for the area where you want to calculate the derivative to make the functions match the requirements.

mathman said:
The direct proof is simple enough ##H(x)=P(C\le x)=P(A\le x, B\le x)=P(A\le x)P(B\le x)=F(x)G(x)## The last step uses independence of ##A## and ##B##.

Where do you prove the product rule for calculating derivatives here? I think I'm using the fact you posted here in my proof.

etotheipi said:
I wondered how you deduced ##h(x)## without going via. the CDF first? Usually to derive the PDF of the ##\text{max}## function I would have thought you say, assuming independence, ##H(x) = P(C \leq x) = P(A \leq x) P(B \leq x) = F(x)G(x)## and then make use of the product rule to find ##h'(x)##.

I think the point here is basically I calculate h by being clever. There pdf can be interpreted as in a small area ##\Delta x## around ##x##, the probability ##A## is in that region is ##f(x)\Delta x##. Similar for ##B## and ##g(x)##. Then if ##C## is in that ##\Delta x## region, the probability it's because ##A## is in that region and ##B##is smaller is ##G(x)f(x) \Delta x##. And a similar formula for the other way around. Adding them up gives the probability that ##C## is in the region. There's a small issue where I have double counted some of the times where both ##A## and ##B## are both in the region, but the chance of that is proportional to ##(\Delta x)^2## so goes to zero fast enough it doesn't contribute to the probability density function.

etotheipi
Office_Shredder said:
I think the point here is basically I calculate h by being clever. There pdf can be interpreted as in a small area ##\Delta x## around ##x##, the probability ##A## is in that region is ##f(x)\Delta x##. Similar for ##B## and ##g(x)##. Then if ##C## is in that ##\Delta x## region, the probability it's because ##A## is in that region and ##B##is smaller is ##G(x)f(x) \Delta x##. And a similar formula for the other way around. Adding them up gives the probability that ##C## is in the region. There's a small issue where I have double counted some of the times where both ##A## and ##B## are both in the region, but the chance of that is proportional to ##(\Delta x)^2## so goes to zero fast enough it doesn't contribute to the probability density function.

I think this is pretty similar to a standard intuitive argument for the product rule (not using probabilistic language). Consider a rectangle with side lengths ##f(x)## and ##g(x)##, so its area is ##f(x)g(x).## Change ##x## by a small amount ##\Delta x## and apply the same reasoning.

etotheipi
Yep, totally agree with that.

Office_Shredder said:
I think the point here is basically I calculate h by being clever. There pdf can be interpreted as in a small area ##\Delta x## around ##x##, the probability ##A## is in that region is ##f(x)\Delta x##. Similar for ##B## and ##g(x)##. Then if ##C## is in that ##\Delta x## region, the probability it's because ##A## is in that region and ##B##is smaller is ##G(x)f(x) \Delta x##. And a similar formula for the other way around. Adding them up gives the probability that ##C## is in the region. There's a small issue where I have double counted some of the times where both ##A## and ##B## are both in the region, but the chance of that is proportional to ##(\Delta x)^2## so goes to zero fast enough it doesn't contribute to the probability density function.

Cool, thanks. That makes sense.

Infrared said:
I think this is pretty similar to a standard intuitive argument for the product rule (not using probabilistic language). Consider a rectangle with side lengths ##f(x)## and ##g(x)##, so its area is ##f(x)g(x).## Change ##x## by a small amount ##\Delta x## and apply the same reasoning.
Do you mean like\begin{align*} \Delta(f(x)g(x)) = f(x+\Delta x)g(x + \Delta x) - f(x)g(x) &\approx (f(x) + \Delta x f'(x))(g(x) + \Delta x g'(x)) - f(x)g(x)\\ \\&\approx \Delta x \left(f'(x) g(x) + f(x) g'(x) \right) \end{align*}where we dropped the cross term in ##(\Delta x)^2##, so$$\frac{d(f(x)g(x))}{dx} = \lim_{\Delta x \rightarrow 0}\frac{\Delta(f(x)g(x))}{\Delta x} = f'(x) g(x) + f(x) g'(x)$$

Infrared

## What is the product rule in probability?

The product rule in probability is a fundamental concept that states that the probability of two independent events occurring together is equal to the product of their individual probabilities. It can be represented mathematically as P(A and B) = P(A) * P(B).

## How is the product rule used in probability?

The product rule is used to calculate the probability of multiple events occurring together. It is especially useful in situations where the events are independent, meaning that the occurrence of one event does not affect the probability of the other event happening.

## What is an example of using the product rule in probability?

An example of using the product rule in probability is calculating the probability of rolling a 3 and a 5 on two fair dice. Since the events of rolling a 3 and rolling a 5 are independent, we can use the product rule to calculate the probability as P(3 and 5) = P(3) * P(5) = (1/6) * (1/6) = 1/36.

## How does the product rule relate to the multiplication rule in probability?

The product rule is a specific case of the multiplication rule in probability, which states that the probability of multiple events occurring together is equal to the product of their individual probabilities. The product rule is used specifically for independent events, while the multiplication rule can be applied to both independent and dependent events.

## Can the product rule be extended to more than two events?

Yes, the product rule can be extended to any number of independent events. The general formula for calculating the probability of multiple independent events occurring together is P(A and B and C and ...) = P(A) * P(B) * P(C) * ..., where A, B, C, etc. represent the individual events.

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