I was going through the text Introduction to Algorithms by Cormen et. al. where I came across the following statement:
If the keys are known to be random real numbers $k$ independently and uniformly distributed in the range $0 \leq k < 1$ , the hash function
$$h(k) = \lfloor km \rfloor$$ satisfies the condition of simple uniform hashing.
Now what I can understand that they are probably considering uniform disturbution in the "continuous" sense and not in the discrete sense. Had it been in the discrete sense then suppose for $n$ keys the probabity mass function (p.m.f) shall be constant and equal to $1/n$ and so it shall be equally likely for each key to be used in the hashing there-by yeilding the desired result.
But we seem sort of in trouble if the distribution being referred to is continuous (I feel so because of the line: "uniformly distributed in the range $0 \leq k < 1$")
Let $f(x)$ be the associated probability density function (p.d.f) and from the given information we have $f(x)=1$,(which is quite easily found, integrating $f(x)$ in the range $0$ to $1$ and equating it with $1$ and noting that in uniform distribution the p.d.f is a constant).
Now though the p.d.f is a constant but p.d.f is not the probability. Rather probability at a spectrum point is $0$. Now how to use this result to get to the claim of the authors.
Or am I entirely at fault considering the distribution to be continuous?
(There is an answer here, but it does not go into this detail as the question there is different after all).