Doubt in CDF and memoryless property I am a beginner. I recently started probability and statistics from Ross. 
 Kindly clarify me the following
Let $X$ be an exponential distributed random variable with parameter 3. Suppose $Y = F(X)$ where $F$ is the cdf of $X$. Then, Y is given as uniform distribution on interval $[0,1]$
My doubt --> $F(x) = 1-e^{-3x}$ -->how is this uniform distribution. Uniform distribution means (from what i understood) in an interval pdf(prob density function) must be constant and area = 1. But graph of $F(x) = 1-e^{-3x}$ (exponential after mirrored with respect to x axis first, then y-axis mirrored and then translted along y firection by +1. (see here for a plot)
this is not uniform right???
2) I understood mathematically memoryless property. But how to vizualise it?$ P(X>t+s|X>s)=P(X>t) $
For ex.,in case of exponential distibution, $P(X>t+s|X>s)$ graph why it shifts to $X>t$ to the right? Kindly enlighten
 A: 1) $Y=F(X)$ is a transformation applied on $X$, just like $Y=X^2$ or $Y=\sqrt{X}$. So, $F(X)$ is not the PDF of $Y$, it's a transformation applied on $X$. The author chooses this transformation function to be the CDF of $X$ (for some reason of course, but that's not the question here). So, that is not the PDF of $Y$ (if it is, where is $y$ in it?).
We just need to find the PDF of $Y$. A typical method is to first finding the CDF and then differentiating. From now on, using subscripts in CDF/PDFs to prevent any confusion:
$$\begin{align}F_Y(y)&=P(Y\leq y)=P(F_X(X)\leq y)\\&=P(X\leq F^{-1}_X(y) )\\&=F_X(F_X^{-1}(y))\\&=y\end{align}$$
And, $$\frac{d F_Y(y)}{dy}=1=f_Y(y)$$
note that since $Y=F_X(X)$ can take values in $[0,1]$, PDF is $1$ in $[0,1]$ and $0$ elsewhere.
Also note that, we haven't used anything related to exponential distribution; because this is a generic method. 
2) Graph/plot of $P(X>t+s|X>s)$ has $t$ as x-axis variable and doesn't shift right, but PDF $f_{X|X>s}(x)$ does, and starts from $s$ instead of $0$ because if we know that $X>s$, there is no density mass left to the left of $s$.
