Expressing the probability density function of $Ax$ in terms of the pdf of $x$

Question

I understand that, for example, you might have a density function which measures the probability of observing an outcome in a certain interval measured in feet, but someone wishes to use meters instead of feet. The density function will have the same shape, but the argument will be scaled by the conversion factor between feet and meters. I know also that the shape of a scale invariant function is indistinguishable if the units on the x and y axis are scaled.I also completely understand the derivation of $f(Ax)=\frac{1}Af(x)$ which can be found by clicking the link below. Finally, I know that $f(Ax)=\frac{1}Af(x)$ is only valid for Probability Density Functions which I shall denote by $f(x)$ What i'm after is an answer to the following question.

My question is; after rearranging the scale invariance property: $f(Ax)=\frac{1}Af(x)$ gives $Af(Ax)=f(x)$. I'm now going to attempt to prove that $f(x)$ satisfies $f(Ax)=\frac{1}Af(x)$ for two cases: $f(x)=$ a constant & $f(x)=$ a power of $x$.

Case 1: Letting $f(x)=\frac{1}6$, as $\frac{1}6$ is a valid probability density function on the domain $(1,7)$ and satisfies $$\int_{1}^{7}f(x)dx=1$$ Now I know that constant $A$ scales the argument. Now since the equation $Af(Ax)=f(x)$ is true for all $A$; as it's just a scaling factor. So letting $A=4$ for example. This turns $Af(Ax)=f(x)$ into $4*\frac{1}6=\frac{1}6$, which is obviously not true but $WHY$ not? Can you give me a numerical example of when $f(Ax)=\frac{1}Af(x)$ is satisfied? Does $f(Ax)=\frac{1}Af(x)$ satisfy all $x$?

By the way; i'm aware that $f(4x)$ could mean a stretch of scale factor $\frac{1}4$ parallel to $x$-axis might turn $Af(Ax)=f(x)$ into $4*\frac{1}4*\frac{1}6=\frac{1}6$ and have rejected it as an answer since $f(x)=\frac{1}6$ has domain $1\leq\ x\leq7$, but the function $f(4x)$ on the other hand, is then only defined when $1\leq\ 4x\leq7$, or $\frac{1}4\leq\ x\leq\frac{7}4$, so the two functions $f(x)$ and $f(4x)$ don't have the same domain, so $f(4x)$ cannot be a constant multiple of $f(x)$.

Case 2: Letting $f(x)=\frac{3}7x^2$, as $\frac{3}7x^2$ is a valid probability density function on the domain $(1,2)$ and satisfies $$\int_{1}^{2}f(x)dx=1$$ Again letting $A=4$, this turns $Af(Ax)=f(x)$ into $4*\frac{3}7(4x)^2=\frac{3}7x^2$, which is obviously not true again. Similarly, if $f(4x)$ could mean a stretch of scale factor $\frac{1}4$ parallel to $x$-axis might turn $Af(Ax)=f(x)$ into $4*\frac{3}7(\frac{1}4x)^2=\frac{3}7x^2$ which is clearly false.

If everything I've said so far is right then I've just proven that there is no Probability Density Function $f(x)$ that satisfies the Scale Invariant Property for Probability Density Functions: $f(Ax)=\frac{1}Af(x)$. Since no $f(x)$ satisfies that formula looks like I've just proven the formula is wrong (even though the derivation is correct - see http://nrich.maths.org/5937/solution). What am I missing here?

Thanks in advance.

This URL: http://nrich.maths.org/5937 is for the NRICH Cambridge Mathematics site, and it will take you to the full question and answer of which this post is about.

Best Regards

Answer 1

The problem here is that you are using $f$ to denote the pdf of $X$ as well as the pdf of $AX$. However, the pdfs two random variables $X$ and $AX$ are different functions.

Let $f_X(x)$ denote the pdf of $X$ and let $f_Y(y)$ denote the pdf of $Y = AX$.

The cdf of $X$ is $F_{X}(x) = \Pr[X \le x] = \displaystyle\int_{-\infty}^{x}f_X(x)\,dx$.

The cdf of $Y$ is $\displaystyle\int_{-\infty}^{y}f_Y(y)\,dy = F_Y(y) = \Pr[Y \le y] = \Pr[AX \le y] = \Pr[X \le \tfrac{y}{A}] = F_X(\tfrac{y}{A})$.

Now, differentiate both sides to get $f_Y(y) = \dfrac{1}{A}f_X(\tfrac{y}{A})$.

To get this in the form you have, substitute $y = Ax$ to get $Af_Y(Ax) = f_X(x)$.

Answer 2

I think that the confusion arises by the fact that $f(x)$ must be a probability density function. Thus $f(x)\geqslant 0$ and \begin{equation*} \int dxf(x)=1. \end{equation*} Let now \begin{equation*} f(Ax)=kf(x). \end{equation*} Now \begin{equation*} \int dxf(Ax)=\frac{1}{A}\int dxf(x)=\frac{1}{A} \end{equation*} and \begin{equation*} \int dxf(Ax)=k\int dxf(x)=k. \end{equation*} Thus the scaled function can only be a probability density function if $k=1/A $. Thus this requirement fixes $k$.

Answer 3

By the way; i'm aware that $f(4x)$ could mean a stretch of scale factor $\frac{1}4$ parallel to $x$-axis might turn $Af(Ax)=f(x)$ into $4*\frac{1}4*\frac{1}6=\frac{1}6$ and have rejected it as an answer since $f(x)=\frac{1}6$ has domain $1\leq\ x\leq7$, but the function $f(4x)$ on the other hand, is then only defined when $1\leq\ 4x\leq7$, or $\frac{1}4\leq\ x\leq\frac{7}4$, so the two functions $f(x)$ and $f(4x)$ don't have the same domain, so $f(4x)$ cannot be a constant multiple of $f(x)$.

In the above argument you missed the fact that the domain of the new probability density function also scales accordingly. Suppose you were measuring something in meters and you had probability of the measurement distributed within domain 1m to 7m, then if you change your measurement system to centimeters then the domain will change to [100cm,700cm]. As the other answers suggest the confusion is from using the same symbol for both the functions (before scaling and after scaling).

Case 2 Going by the advice from other answers we should denote different probability function by different symbols. Let $y = Ax$ where $x$ is a random variable with probability density $f_X(x) = \frac{3}{7}x^2$ and domain $[1,2]$. Using the result of scale invariance, the probability density of $y$ is $f_Y(y) = \frac{1}{A}f_X(x)=\frac{1}{A}f_X(\frac{y}{A})$ with domain $[A, 2A]$. Using $A=4$, we have $f_Y(y) = \frac{1}{4}(\frac{3}{7}(\frac{y}{4})^2)$. Now, we can check the equality at any point $x=\frac{3}{2} \in [1, 2]$. $f_X(x) = \frac{3}{7}*\frac{9}{4}$ while $f_Y(4*\frac{3}{2}) = \frac{1}{4}*\frac{3}{7}*\frac{9}{4}$ . For the computed values $4*f_Y(4*\frac{3}{2}) = f_X(\frac{3}{2})$ is true.