# Find PMF of X^2 if X~Dunif(0,1,…,n)

(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

1. Find CDF given X is DUnif:

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

1. 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?

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.