Proving the product rule using probability

[Office_Shredder]

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.

 

[jedishrfu]

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

 

[mathman]

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 $B$.

 

[etotheipi]

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 !

 

[Office_Shredder]

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.

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.

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.

 

[Infrared]

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.

 

[Office_Shredder]

Yep, totally agree with that.

 

[etotheipi]

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.

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$$\frac{d(f(x)g(x))}{dx} = \lim_{\Delta x \rightarrow 0}\frac{\Delta(f(x)g(x))}{\Delta x} = f'(x) g(x) + f(x) g'(x)$$