# probability distribution of the maximum

Let T be a random variable giving the time to failure of led lights that follow exponential distribution with a mean value of 15 000 hours.

We put three new lights at the same time. Find the cumulative distribution of $T_{max}$ and then find its probability density function.

I assume the cumulative distribution will be a product of $(1-e^{(-1/15000)t_1}) (1-e^{(-1/15000)t_2})(1-e^{(-1/15000)t_3})$ but I am not sure how to get the probability density function. Any hints?

• (i) You really don't know how to get a pdf from a cdf? (ii) your cdf looks wrong in a couple of ways. Can you start with the cdf for a single light and then explain how you got that particular cdf for the maximum? – Glen_b -Reinstate Monica Mar 13 '15 at 2:18
• the pdf is the first derivative of the cdf but it doesn't get me the answer. I am totally lost. – user218698 Mar 13 '15 at 2:21
• Here's a hint: forget you've seen the answer and focus on figuring out how to do the question. Giving people the answers seems to induce a mentality of "let's press random buttons until I get something that's the same as that and then stop" instead of careful thinking about what's needed to solve the problem. You'd be strongly advised to get into the habit of acting as if there were never answers supplied. Write your answers in such a way that skeptical people would be convinced that the answer must be correct. – Glen_b -Reinstate Monica Mar 13 '15 at 2:24
• In short, edit your question to show your reasoning. Justify every step. Apply reasonableness checks every time you can. Your answer doesn't pass several reasonableness checks so you don't need to see an answer to know it's wrong. – Glen_b -Reinstate Monica Mar 13 '15 at 2:25
• So to repeat part of my initial comment "Can you start with the cdf for a single light ..." – Glen_b -Reinstate Monica Mar 13 '15 at 2:31

It's important to start off with properly defined variables and events, and to develop the calculations carefully, so you can see where the mistakes are (don't jump steps, every time you did, you had mistakes that were then hard to spot). Something like this:

Let $T_1$ be the time until failure for light 1, and similarly define $T_2$ and $T_3$.

Let $W$ be the time until all three lights fail.

The event $W> w$ is equivalent to $(T_1 > w)\cap(T_2 > w)\cap(T_3 > w)$. Hence

$\:P(W > w)=P((T_1 > w)\cap(T_2 > w)\cap(T_3 > w))$

$\qquad\qquad\quad=P(T_1 > w)\,\cdot\,P(T_2 > w)\,\cdot\,P(T_3 > w)$ (independence).

Now $P(T_1>w) = 1-P(T_1\leq w)=...$ et c.

Then work out $P(W\leq w)$ from that.

When you can do it like this without any risk of an error, there are some shortcuts you can get away with ... but to be honest, I tend to use few shortcuts on problems of this sort. In the end, it's faster to do it properly the first time than to redo it three (or more) times.

(Nevertheless, I managed to make a mistake here; I originally had $\leq$ where I should have had $>$. It was, however, easier to correct because what was there was more explicit.)

• As we have $i=1,2,3$ should we have: $$P(T_i <= w)=(1-e^{-t/θ})^3$$ – user218698 Mar 13 '15 at 12:59
• No, that can't be right, no matter which value of $i$ you pick. You seem to be confused about what $P(T_i\leq w)$ means. Think about a specific value for $i$, and put any other dummy (perhaps $t$ say) in place of $w$ while you think about what that means in words. Then put $w$ back and note that the only thing that changes is the dummy variable itself. I'll amend my answer to be more concrete. – Glen_b -Reinstate Monica Mar 13 '15 at 13:40
• For me $P(T_i≤w)$ means $P(T_1≤w,T_2≤w,T_3≤w)=(1-e^{-t/θ})^3$. Am I so far from the truth? – user218698 Mar 13 '15 at 14:18
• The problem is that $P(T_i\leq w)$ doesn't mean $P(T_1≤w,T_2≤w,T_3≤w)$. The second statement would be correct, if you have defined that $\theta=15000$. – Glen_b -Reinstate Monica Mar 13 '15 at 14:22