Find $c$ for $P(\frac{5}{12}-c \le Y \le \frac{5}{12}+c)=\frac{1}{2}$

Question

Find $c$ for $P(\frac{5}{12}-c \le Y \le \frac{5}{12}+c)=\frac{1}{2}$

We also have that $$f_Y(y) = k\sum_{i=0}^{\infty}y^i, y\in(1/3, 1/2) \\ \implies k\int_{1/3}^{1/2}\frac{1}{1-y}dy \implies k=\frac{1}{\log(\frac{4}{3})}$$

What I have tried:

I had thought to apply chebyshev's inequality, in this case $\sigma=1$, so we have

$P(|Y-\frac{5}{12}|\le c)=\frac{1}{2} \\ \frac{1}{2} = 1-\frac{1}{c^2} \implies c = \sqrt{2} \\ \implies P(|Y-\frac{5}{12}| \le \sqrt{2})=\frac{1}{2}$

However, a part of me thinks I need to use the standard normal distribution inequality to figure this out instead.

Something like $$P\left(\frac{\frac{5}{12}-c}{1} \le Y \le \frac{\frac{5}{12}+c}{1} \right)=\frac{1}{2}$$

and so we have to show that

$$\Phi(0) = \frac{1}{2} \\ \Phi(\frac{5}{12}+c)-\Phi(\frac{5}{12}-c)=\frac{1}{12} \\ \implies \Phi(\frac{5}{12}+c)-\left\{1-\Phi(c-\frac{5}{12}) \right\}\\ \text{we did that }\Phi(-z)=1-\Phi(z) \\ \implies \Phi(\frac{5}{12}+c) + \Phi(c-\frac{5}{12})-1=\frac{1}{2}$$

If I plug in the value of $\sqrt{2}$ like in the previous inequality, then I do not get the value of $\frac{1}{2}$. How do I approach this correctly with the chebyshev or standard normal?

Answer 1

If the distribution is not "normal" then why would you think of applying tricks from normal distribution? . Secondly it is Chebycheff's Inequality. So applying it won't yield you any "equality" for $c$.

$$P(\frac{5}{12}-c<Y<\frac{5}{12}+c)=k\int_{\frac{5}{12}-c}^{\frac{5}{12}+c}\frac{1}{1-y}\,dy = k\ln\bigg(\frac{\frac{7}{12}+c}{\frac{7}{12}-c}\bigg)$$.

This means that $$k\ln\bigg(\frac{\frac{7}{12}+c}{\frac{7}{12}-c}\bigg)=\frac{1}{2}$$

So $$\bigg(\frac{\frac{7}{12}+c}{\frac{7}{12}-c}\bigg)=e^{\frac{1}{2k}}$$.

Now solve for $c$.

It would be wise of you to go back a step and re-read the definitions of the Probability Density function and why integrating over a set gives the probability of the random variable lying in that set.