Find PMF of X^2 if X~Dunif(0,1,...,n)

Question

Follow up on this: Find PMF of $X^2$ if $X$~Dunif

(I do not have enough "reputation points" to comment, so if this is an inappropriate way to ask for a follow up, please let me know)

Is this a correct way to solve:

$Y=X^2$

X~DUnif(0,1,...,n)

1. Find the PMF relating X and y:

$F_Y(y) = P(Y<y) = P(X^2<y) = P(X<\sqrt{y})$

2. Find CDF given X is DUnif:

$F_X(\sqrt{(y)}) = \int_0^{\sqrt{y}}\frac{1}{n}dx = \frac{\sqrt{y}}{n}$

3. Take derivative to find PDF:

$\frac{1}{2n\sqrt{y}}$

for y in {0,1,...,n}

I think I am missing something due to the squared/square root. Any help?

Answer 1

If $X$ is uniformly distributed over $\{0,1,\ldots,n\}$, then the distribution of $Y=X^2$ is straightforward to determine: \begin{align} \mathbb P(Y=1) &= \mathbb P(X=0)+\mathbb P(X=1)=\frac 2{n+1}\\ \mathbb P(Y=k^2) &= \mathbb P(X=k) = \frac1{n+1},\ k=2,\ldots,n. \end{align} The distribution function of $Y$ is thus $$ F_Y(y) = \frac2{n+1}\mathsf 1_{[1,\infty)}(y) + \sum_{k=2}^n\frac1{n+1}\mathsf 1_{[k^2,\infty)}(y). $$ Taking a derivative does not make sense here since the random variables are not continuous.