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I have encountered this problem in the book Probability and Random Processes by Grimmett and Stirzaker. The problem is:

Let $X_1, X_2, X_3$ be independent (discrete) random variables taking values in the positive integers and having mass functions given by $\mathbb{P}(X_i=x)=(1-p_i)p_i^{x-1}$ for $x=1, 2, ....$ and $i=1,2,3$.

Show that $\mathbb{P}(X_1<X_2<X_3)=\dfrac{(1-p_1)(1-p_2)p_2p_3^3}{(1-p_2p_3)(1-p_1p_2p_3)}$

I actually find this kind of problems too vague, because I know how to calculate $\mathbb{P}(X\leq x)$ but I can't calculate the probability of (in)equality of different random variables $X, Y$ like $\mathbb{P}(X<Y)$. Is there a method for that?

What I tried so far is to write it down as $\mathbb{P}(X_1<X_2<X_3)=\mathbb{P}(X_1<X_2, X_2<X_3)=\mathbb{P}(X_1<X_2)\mathbb{P}(X_2<X_3)$. But then I don't know how to go further...

I apperciate your help and hints. Thanks!

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  • $\begingroup$ not sure your last expression is correct. I would try $$\mathbb{P}[X_1<X_2<X_3] = \mathbb{P}\left[\left. X_1 < X_2 \right| X_2 < X_3 \right] \cdot \mathbb{P}[X_2 <X_3]...$$ $\endgroup$
    – gt6989b
    Commented Nov 28, 2016 at 18:31
  • $\begingroup$ I still have no clue how to continue.. :( $\endgroup$
    – Shashi
    Commented Nov 28, 2016 at 18:51

2 Answers 2

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We use this law https://en.wikipedia.org/wiki/Law_of_total_probability, and the fact that the variables are mutually independent

$P(X_1<X_2<X_3)=\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{P(X_1<X_2<X_3|X_2=j,X_1=i)}P(X_2=j,X_1=i)}$

$P(X_1<X_2<X_3)=\sum_{i=1}^{\infty}{\sum_{j=i+1}^{\infty}{P(j<X_3|X_2=j,X_1=i)}P(X_2=j)P(X_1=i)}$

By independence, $P(j<X_3|X_2=j,X_1=i)=P(X_3>j)$, and also by the law of total probability, $P(X_3>j)=\sum_{k=j+1}^{\infty}{P(X_3=k)}$

So we have $P(X_1<X_2<X_3)=\sum_{i=1}^{\infty}{\sum_{j=i+1}^{\infty}{\sum_{k=j+1}^{\infty}{P(X_3=k)}}P(X_2=j)P(X_1=i)}$

$=(1-p_1)(1-p_2)(1-p_3)\sum_{i=1}^{\infty}p_{1}^{}{\sum_{j=i+1}^{\infty}p_{2}^{j-1}{\sum_{k=j+1}^{\infty}{p_{3}^{k-1}}}}$

Using the fact that $\sum_{j=i+1}^{\infty}{u^{j-1}}=\frac{u^{i}}{1-u}$, you can conclude

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  • $\begingroup$ This helps me a lot! Thanks! $\endgroup$
    – Shashi
    Commented Nov 28, 2016 at 19:19
  • $\begingroup$ Thanks Byron Schmuland, I fixed it $\endgroup$
    – Canardini
    Commented Nov 28, 2016 at 19:21
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For two random variables the answer is: \begin{eqnarray*} \mathbb{P}(X_2<X_3)&=&\sum_{1\leq j<k} (1-p_2)(1-p_3)p^{j-1}_2 p_3^{k-1}\\[8pt] &=&\sum_{1\leq j}(1-p_2)(p_2p_3)^{j-1}\\[8pt] &=&{1-p_2\over 1-p_2p_3}. \end{eqnarray*} I will leave the rest to you.

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  • $\begingroup$ This is very similar to the solution given for the exercise. But I cant figure out how to go from the first equality to the second. Maybe it is because I'm not familiar with the notation in the summation. $\endgroup$ Commented Sep 1, 2023 at 23:49
  • $\begingroup$ Nvm, this question is exactly about this: math.stackexchange.com/questions/536055/… $\endgroup$ Commented Sep 1, 2023 at 23:53

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