Find C value given CDF

Question

Given $Y$ a continuous random variable which takes no values with positive probability. Let $F$ be its CDF given by the table:

$F(y)=\left\{\begin{matrix} y<0 & C \log (1-y)/y\\ y>=0 & 1 \end{matrix}\right.$

Find the value of C.

I tried to get the probability i.e. $P(Y=0)$ from the corresponding CDF. $P(Y>=y)=1-P(Y<y)=1-C \log (1-y)/y$

$P(Y<y)=C \log (1-y)/y$

Thus $P(Y=y)=P(Y>=y)-P(Y<y)$

Equating this to $0$ and substituting the value of $y$ as $0$, I am finding the value of $C$.

But the answer is not coming correct. Can you please let me know if the approach to the above formulation is correct and at which step it is going wrong?

Answer 1

By assumption, $F$ is continuous at $0$. The right limit is $1$, while the left limit is $$\lim_{y\to 0^-}{\frac{C\log(1-y)}{y}}=C\lim_{y\to 0^-}{\frac{\log(1-y)}{y}}=C\cdot(-1)=-C.$$ Therefore, $-C=1$, which gives $C=-1$.

Besides, in your assumption, $Y$ takes no values with positive probability. There is no wonder why $P(Y=0)=0$.