There are LED light bulbs such that the probability that a bulb still works after x months of continuous use is equal to $\frac{1}{(1+x)^2}$. What is the PDF for the lifetime of such a bulb? What is the expected lifetime and the median lifetime?
They're giving me the probability that a bulb still works after X months. That is the PDF, right? I think that makes sense, because it satisfies the properties: $\int_D f(x)=1$, and $f(x) > 0, \forall{x}$, but the wording is making me uneasy.
$\mathbb{E}[X]= \int_D xf(x)$, and the median is just $\int_0^m f(x) = \frac 1 2$, I believe. I'm just having trouble deciphering if I have f(x) right, with f(x) being the same as the given equation.
Best Answer
The probability that the light bulb will survive more than $x$ months is $ \frac{1}{(x+1)^2}$.
In other words, $P(X \gt x) = \frac{1}{(x+1)^2}$
So, $ \text { CDF } ~ F_X(x) = P(X \lt x) = 1 - P(X \gt x)$
$ \displaystyle = 1 - \frac{1}{(x+1)^2}$
$ \therefore \text { PDF } ~ \displaystyle f_X(x) = \frac{2}{(x+1)^3}, 0 \lt x \lt \infty$
Can you take it from here?