Multiplying a Probability distribution function by a constant This is a pretty basic question, I understand, but in my head there is something that doesn't make sense.
Let's say I have a uniform probability density function X that goes from [0,6]. The pdf would thus be 1/(6-0), which is just 1/6.
Now, let's say I multiply this pdf X by 2, to get Y= 2X. Would the pdf then be 2*(1/6)? If so, wouldn't this break the condition that the integral of a pdf has to be equal to 1? What would I do? And for whatever answer I get, would that work for all pdf's (binomial, Poisson, uniform, normal, etc).
 A: You can explain it in this way:
$$X \sim Uni(0,6) \Rightarrow F_X(t)=P(X\leq t) = \frac{t}{6} \\
Y = 2X \Rightarrow F_Y(s) = P(Y\leq s) = P(2X \leq s ) = P (X\leq \frac {s}{2})
=_{s\in(0,12)} \frac{s}{12}$$
Hence it is similar to what you thought, but opposite, we got that the density function of $Y$ is $\frac{1}{12}$ which is half, and the explanation of this is in the computation above
The same explanation goes for every other distribution, BUT that doesn't mean you will get the property of anti-homogeneous we got here.
For example, if $X ~ \exp(1)$ then $f_X(x)=e^{-x}$ and for $Y=2X$ we get
$$ F_Y(t) = P(Y\leq t) = P(X \leq \frac{t}{2}) = 1-e^{-\frac{t}{2}}$$
Hence, in this case $f_Y(y) = e^{-\frac{y}{2}}$ which means $Y\sim\exp(\frac{1}{2})$ and NOT $(\frac{1}{2}\exp{1})$ as you could (maybe) have expected
A: Using fundamental transformation  theorem you get
$$f_Y(y)=f_X\left(g^{-1}(y)\right)|\frac{d}{dy} g^{-1}(y)|  $$
$$f_Y(y)=\frac{1}{6}\cdot\frac{1}{2}\cdot\mathbb{1}_{(0;12)}(y)$$
Being
$$y=2x\rightarrow x=\frac{y}{2}\rightarrow x'=1/2$$
In other word, $Y\sim U(0;12)$
Important observation: in your case you are not multiplying the distribution by 2 but THE RANDOM VARIABLE. It is not the same.
