Finding K for a cumulative distribution function.

Question

I have the following CDF for some fixed number $k$:

When $x$ is smaller than $1$ then $F(x) = 0$
When $x$ greater or equal to $1$ then $k - \frac{1}{⌊x⌋}$ applies.

However, I cannot figure out how to find $k$. I believe that $k$ cannot be larger than $1$ because if $x = 1$ then $2-\frac{1}{1} = 1$

Also, I am told that $FX(∞)=1$ is the key to find $k$.

score 1 · Accepted Answer · answered Aug 01 '19 at 11:48

1

As $x \to \infty$ $[x]$ also tends to $\infty$ so $F(x) \to k-0=k$. But this must be $1$ for a CDF so the answer is $k=1$.

answered Aug 01 '19 at 11:48

Kavi Rama Murthy

Would this scenario also be plausible: $F(1) = 1 - 1 = 0$ ? – Libor Zachoval Aug 01 '19 at 11:56
No. You cannot say that $F(1)$ must be $0$ simply because $F(x)=0$ for $x<1$. CDF's may not be continuous. – Kavi Rama Murthy Aug 01 '19 at 12:01
So in that case, where did the $0$ come from in here "$F(x)→k−0=k$"? – Libor Zachoval Aug 01 '19 at 12:08
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If $x >N$ then $n \leq x <n+1$ for some $n \geq N$. Hence $[x]=n \geq N$. This shows that $[x] \to \infty$ as $x \to \infty$. Hence $\frac 1 {[x]} \to 0$ as $x \to \infty$. – Kavi Rama Murthy Aug 01 '19 at 12:11
Ah yes, I understand now. Thank you – Libor Zachoval Aug 01 '19 at 12:11
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More simply $[x] >x-1$ always, so $[x] \to \infty$ as $ x \to \infty$. – Kavi Rama Murthy Aug 01 '19 at 12:12

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