Vrftsjtryk

$T_1, ... , T_n$ time of life of instruments with geometric distribution of parameter p independents.
I define S as the first n instants of failure and U as the last n instant of failure.
I want to find the law of S and the density of U.

This is what I did so far:

To find the law of S I start by calculating:
$$mathbb{P}(S = k) = mathbb{P}(S geq k) - mathbb{P}(S geq k + 1) $$
$$mathbb{P}(S = k) = mathbb{P}(S < k+1) - mathbb{P}(S < k)$$
But here I am stuck on what I should do because I still do not know the distribution.

I do the same reasoning with U as follows:
$$ mathbb{U}(S = k) = mathbb{U}(U leq k) - mathbb{U}(S leq k + 1) $$
But I still get nowhere.

Any suggestions?

EDIT:

Following the suggestions I managed to say that:

$$mathbb{P}(S=k) = [mathbb{P}(T_1 geq k)]^n = [1 -mathbb{P}(T_1 < k)]^n = $$
$$ = [1 - [1-(1-p)^k]^n = [(1-p)^k]^n $$

Now the exercise asks me to identify random variable, saying it is a geometric distribution with parameter $1-(1-p)^n$. However I am struggling to see how this is true.

Moreover, how should I study U? Is it the maximum instead of the minimum? I would love to see a hint of solution using my procedure as well.

First of all defining $S$ as the "first n instants [sic] of failure" makes little sense, as you have only been given $n$ lifetimes, so for there to be $n$ instances of failure would require all $n$ lifetimes to terminate. This confusion is likely why Drhab commented as such, suggesting instead to look at $S=min{T_1,dotsc,T_n}$ and provided the general formula for all minimums of IIDRVs: $P(S> k)=P(T_1> k)^n$. This is interpreted as the first instance of failure out of the $n$ lifetimes, i.e. the minimum lifetime.

Then, with $P(T_1leq k)=1-(1-p)^k$ (assuming CDF corresponding to when the $T_i$ take values in $mathbb{N}$), we first have $P(T_1>k)=(1-p)^k=:q^k$ where $q=1-p$ for ease of notation. Then substituting, we have $P(S>k)=(q^k)^n=(q^n)^k$, so that indeed $P(Sleq k)=1-(q^n)^k$ is the CDF of a Geometric RV with new parameter $tilde{p}=1-q^n=1-(1-p)^n$.

How can we tackle $U$? Well, let's see how that minimum formula was derived in the first place and hope that informs us on a similar manner to derive a formula for the maximum of IIDRVs.

If the minimum of $n$ variables is greater than a number $k$, then all of those variables must be greater than $k$, i.e the following set equalities hold
$${min{T_1,dotsc, T_n} >k}={T_1 >k,dotsc, T_n>k}.$$
That's just from the definition of minimum. Thus, taking probabilities and writing $S=min{S_1,dotsc, S_n}$, we have
$$P(S >k)=P(T_1>k, dotsc, T_n>k),$$
and finally the IID assumption greatly simplifies the RHS further, to
$$P(S>k)=P(T_1>K)P(T_2>k)cdot dotsc cdot P(T_n >k),$$
first using independence, then using the identical distributions, we finally get
$$P(S>k)=P(T_1>k)^n.$$

Can you now, after seeing this argument for minimum, derive the analogous formula for maximums? Please comment for further clarification.

EDIT
If $U=max{T_1,dots, T_n}$, then we know that $P(Uleq k)=P(T_1 leq k)^n=(1-q^{k})^n$ by definition of maximum and since the $T_i$ are IID. To find $mathbb{E}(U)$ we use the alternative formula for expectations for discrete non-negative RVs:
$$E(U)=sum_{k=1}^infty P(Ugeq k).$$
Since $P(Ugeq k)=1-P(Uleq k-1)=1-(1-q^{k-1})^n,$ we must compute the series,
$$sum_{k=1}^infty 1-(1-q^{k-1})^n.$$
In principle this series can always be computed by using the binomial theorem to expand $(1-q^{k-1})^n$ and then use theory of geometric series to compute the resulting series' in closed form. For $n=2$, we have
$$mathbb{E}(U)=frac{3-2p}{p(2-p)}approx 2.667$$
when $p=0.5$, for example.