Exponential distribution - inequality

[mathmari]

Hey!

We consider the exponential distribution.

I want to show that $$\mathbb{P}\left (\left |X-\frac{1}{\lambda}\right |\leq \lambda \right )\geq \frac{\lambda^4-1}{\lambda^4}$$

I have shown so far that \begin{align*}\mathbb{P}\left (\left |X-\frac{1}{\lambda}\right |\leq \lambda \right )&=\mathbb{P}\left (\frac{1-\lambda^2}{\lambda} \leq X\leq \frac{1+\lambda^2}{\lambda}\right ) \\ & =F\left (\frac{1+\lambda^2}{\lambda}\right )-F\left (\frac{1-\lambda^2}{\lambda}\right )=e^{-1+\lambda^2}-e^{-1-\lambda^2}\end{align*} Is this correct? How could we continue? (Wondering)

 

[mathmari]

Now I have an other idea. Do we maybe use Chebyshev inequality ?

We have that $$\mathbb{P}\left (\left |X-\frac{1}{\lambda}\right |< \lambda \right )=1-\mathbb{P}\left (\left |X-\frac{1}{\lambda}\right |\geq \lambda \right )\geq 1-\frac{\text{Var}(X)}{\lambda^2}=1-\frac{\frac{1}{\lambda^2}}{\lambda^2}=1-\frac{1}{\lambda^4}=\frac{\lambda^4-1}{\lambda^4}$$

But now we have $$\mathbb{P}\left (\left |X-\frac{1}{\lambda}\right |< \lambda \right )$$ instead of $$\mathbb{P}\left (\left |X-\frac{1}{\lambda}\right |\leq \lambda \right )$$ (Wondering)

 

[S.G. Janssens]

Well done to think of Chebyshev.

The events
\[
\left\{ \left| X - \frac{1}{\lambda} \right| < \lambda \right\} \qquad \text{and} \quad \left\{ \left| X - \frac{1}{\lambda} \right| \le \lambda \right\}
\]
have the same probability because the underlying probability measure is absolutely continuous (with density equal to the exponential density with parameter $\lambda$), so the difference between strict and non-strict inequality does not matter.

 

[I like Serena]

Yep. All good.
And note that in general for a continuous distribution $P(X=x)=0$ so that $P(X>x)=P(X\ge x)$. (Thinking)

 

[mathmari]

Great! Thank you very much! (Happy)