If $X$ is a random variable, and $Y= 2X$, then why isn't it enough to multiply the density function of $X$ by $2$ to find the density function of $Y$?

Question

This may be a dumb question, and I've tried searching online for answers, but I can't seem to wrap my head around it.

So say I have a random variable $X$ and $Y = 2X$. Now I want to find the density function $f_Y(y)$. Why can't I just multiply $f_X(x)$ to find this value. I understand how to derive $f_Y(y)$, by taking the derivative of $F_Y(y) = P(Y \lt y)$, substituting $Y$ for $2X,$ etc. But I don't understand why. What does $Y=2X$ actually mean? Isn't it simply multiplying all the values that $X$ is described by, which is $X$'s density function, by 2? Or is it doing something else entirely?

Again, sorry if this is a very basic question, just can't seem to fully grasp this concept.

Answer 1

A good way of feeling this in your bones is to note the difference between $f(2x)$ and $2f(x)$. We are not at all guaranteed for these to equal one another.

When you consider the CDF $F_Y(y)$ it works that: $$F_Y(y) = P(Y \le y) = P(2X \le y)=P(X\le y/2)=F_X(y/2)$$

A random variable is just a function. It takes things like events and stamps a number on it. If you get into sigma algebras and measure theory you’ll encounter this in detail.

Think of it like stamping numbers on cows in a field. Cowboy X randomly stamps numbers with a certain likelihood, like say having a likelihood of stamping 6’s 40% of the time. Well cowboy Y stamps according to cowboy X, but he always stamps the number twice as large as what X would/did stamp. When X stamps a 6, we know Y stamps a 12. Just the same since the likelihood of X stamping a 6 is 40% we know Y stamping a 12 is also 40%. We’re not inherently changing the structure of the distribution with Y = 2X, we’re just stretching the values out. Just like when you take $f(x)$ and compute $f(2x)$

Answer 2

A density has integral $1$. If $f$ is the density of $X$, $2f$ cannot be a density as it would have integral $2$.

Answer 3

“Why isn't it?” is a strange question when no reason why it is has been suggested. $$ \Pr(2<X<3) = \int_2^3 f(x) \, dx $$ and therefore $$ \Pr(4<2X<6) = \int_2^3 f(x) \, dx = \int_4^6 \cdots\cdots. $$ What function should go where those dots are? Putting $2f$ in place of $f$ would make the first integral above twice as big. But what would it do with any integral from $4$ to $6\text{?}$