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The exercise says:

Let $X$ ~ $G (p)$. Prove that $P (X ≥ k) = (1 - p)^k$, with $k ∈ N$.

My try to demonstrate it:

  • $P (X ≥ k) = 1-P(X
  • $P (X+1 ≥ k+1) = 1-P(X+1
  • $P (X+1 ≥ k+1) = 1-((1-p)^kp)$
  • $P (X+1-1 ≥ k+1-1) = 1-(((1-p)^kp)-1)$
  • $P (X ≥ k) = 1-(((1-p)^kp)-1)$
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    If you think of $X+1$ as the first time in a sequence of coin tosses that the coin lands heads (with probability of landing heads $p$ on each toss), then $X+1\geq k+1$ if and only if the first $k$ tosses landed tails, which happens with probability... – 2017-02-06
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    I am sorry. But I did not quite understand your analogy. – 2017-02-06

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$\bbox[0.5ex]{X\sim\mathcal{Geo}_0(p)}$ means $X$ is a count of failures before a success in an indefinite sequence of independent Bernoulli trials with identical success rate $p$.

$\bbox[0.5ex]{\mathsf P(X\geq k)\quad [k\in\Bbb N]}$ is then the probability for the first success occurring after at least $k$ consecutive failures.   Which is simply the probability for obtaining $k$ consecutive failures.

$$\mathsf P(X\geq k)=(1-p)^k$$

Of course you can show this by using $\bbox[0.5ex]{\mathsf P(X=k)~=~(1-p)^kp}$ (but why is that so)?

$$\begin{align}\mathsf P(X\geq k) &= \sum_{j=k}^\infty (1-p)^j p \\ & = (1-p)^kp\sum_{j=k}^\infty (1-p)^{j-k} \\ & \vdots \end{align}$$

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    The first explanation is the demonstration itself? Or should I do something else? – 2017-02-06
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    @CarlosFrostte The first should suffice but, just in case, you can complete the second. (Hint: It is beginning to look a lot like a Geometric Series.) – 2017-02-07