# Find probability of $X_1 < X_2 < X_3$ where $X_i$ is a random variable

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!

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

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

• This helps me a lot! Thanks! Commented Nov 28, 2016 at 19:19
• Thanks Byron Schmuland, I fixed it Commented Nov 28, 2016 at 19:21

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.

• 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. Commented Sep 1, 2023 at 23:49