# Properties of a cumulative distribution function

Suppose we know that $$F_X(x)=P(X\le x)=x$$. So, if we were to look for $$P(X\le x/n)$$, can we say that $$P(X\le x/n)=\frac{1}{n} P(X\le x)=\frac{1}{n}F_X(x)=\frac{x}{n}$$ given $$n$$ is some constant

• You cannot have $F_X(x)=x$ for all real $x$. Edit the question to make it more precise. Commented Jan 18, 2022 at 9:21
• If $F_X(x)=P(X\le x)=x$ when $0 \le x \le 1$ then for $n\ge 1$ it follows immediately that $P(X\le x/n)=\frac{x}{n}$ when $0 \le x \le n$ since $0 \le \frac x n \le 1$ Commented Jan 18, 2022 at 9:40

No you can't factor out the $$\frac1n$$ in the CDF, the pdf although of the RV $$Y=nX$$ has a density of $$f_Y(y)=\frac d{dy}F_X(y/n)=\frac1n f_X(y/n)$$.
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