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A random experiment has exactly three possible outcomes, referred to as outcomes $1, 2,$ and $3,$ with probabilities $p_1 > 0, p_2 > 0,$ and $p_3 > 0,$ where $p_1 +p_2 +p_3 = 1.$ We consider a sequence of independent trials, at each of which the specified random experiment is performed.) For $i = 1, 2,$ let $N_i$ be the number of trials needed for outcome $i$ to occur, and put $N := N_1 \wedge N_2.$

(a) Show that $N$ is independent of $\{N_1 < N_2\}.$

(b) Evaluate $E[N_1 \mid N_1 < N_2].$

(c) Roll a pair of dice until a total of $6$ or $7$ appears. Given that $6$ appears before $7,$ what is the (conditional) expected number of rolls?

The answer to $b)$ is $1/(p_1+p_2)$ and $c)$ is $3.272727$ but I'm unsure of even where to start for $a)$ or the steps involved in arriving at the answers for $b)$ and $c)$

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  • $\begingroup$ should I somehow use the fact that 𝐸[1𝐸]=𝑃(𝐸) in the proof? $\endgroup$
    – Hannnnn
    Oct 29, 2019 at 13:43

2 Answers 2

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Here is a brute force way to prove (a). I'm sure there is a simpler way to do this, and hopefully someone will post it. We have the joint pdf \begin{align} \mathbb{P}(N_1=k, N_2 = j) &= 1_{k<j} \;p_3^{k-1}p_1(p_1+p_3)^{j-k-1}p_2 + 1_{k>j}\; p_3^{k-1}p_2(p_2+p_3)^{j-k-1}p_1\\ &= p_1p_2p_3^{k-1}\left[ 1_{k<j} \;(p_1+p_3)^{j-k-1} + 1_{k>j}\; (p_2+p_3)^{j-k-1} \right] \end{align} Using the above, one can show that $$ \mathbb{P}(N=n, N_1<N_2 ) = p_3^{n-1}p_1$$ $$ \mathbb{P}(N=n) = p_3^{n-1}(p_1+p_2) $$ $$\mathbb{P}(N_1<N_2 ) = \frac{p_1}{p_1+p_2}$$ And therefore, $$\mathbb{P}(N=n, N_1<N_2 ) = \mathbb{P}(N=n)\cdot \mathbb{P}(N_1<N_2 )$$ proving independence of the events.

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HINT

$N_i$ is the trial at which $i$ first appears. So $N = \min(N_1, N_2)$ is the trial at which $1$ or $2$ first appears. In other words, before the $N$th trial, only $3$ has been appearing. In other words, $N>n$ is the event that the first $n$ trials are all $3$s.

(a) I would directly calculate $P(N>n \mid N_1 < N_2)$ and $P(N > n \mid N_2 < N_1)$ and show that they are equal. Do you see why this implies the required independence?

(b) Consider $P(N_1 > n \mid N_1 < N_2)$. Can you see why this is exponential?

(c) is just an application of (b)

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