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The question is based on an engineering problem and, due to my lack of knowledge in statistics, I found it difficult to express the question clearly in maths language.

The problem is formulated like this: 1) Imagine there is an origin, and a 1D coordinate system (say, L);

2) Along this 1D axis (L), there is a 'rigid' rod attaching to the origin, and its length is a random variable (always positive) and follows a PDF p1.

Question 1: for each given point on the axis L, what is the chance of that point belonging to the thread.

My solution to Q1: for a given coordinate x, for it to 'fall' into the thread, the thread length (a) has to be equal or greater than x. Therefore, the chance of a point x belonging to that rod is equal to the integration of that PDF, from x to infinity.

Can anyone please confirm if my solution above is correct?

If so, the question 2 is as follows: if the near end of the rod no longer attaches to the origin, but the location of it is also a random variable, which follows, say, another, PDF p2. Then, for each given point on the axis L, what is the chance of that point belonging to the rod?

I am so confused by this question 2, as I have no idea what I should do with the potential interplay between these two random events (i.e. the length of rod and the location of near end of the rod). I wonder if anyone here can help me with this, or at least let me know what kind of 'theory' I should look into for solving this problem?

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  • $\begingroup$ Gaussian distributions have no limits, so it would behoove you to specify clearly what you mean by this assumption. Also: Is the thread taut? Is it equally likely to lie to the left or right of the origin? Why do you refer to "interplay between ... random events" when considering just a single random variable? A lot of essential detail seems to be missing. $\endgroup$ – whuber Apr 12 '18 at 20:46
  • $\begingroup$ Hi Whuber, thank you for your reply. Sorry if it wasn't clear as I have little background in statistics and my English is poor. To answer your questions: It is more like a 'rigid' rod and its length is a random variable, so it is not taut. It is only allowed to lie to the right of the origin, so the coordinate system is more like a 'vector'. There are two random variables in Question 2. one is the length of the rod, and another one is the location of the near end of the rod. hope this makes things clearer... $\endgroup$ – Deisler Apr 12 '18 at 20:51
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    $\begingroup$ You don't need any background: just describe the problem you face, in your own terms, as clearly and specifically as possible. Define or explain uncommon technical terms. $\endgroup$ – whuber Apr 12 '18 at 20:52
  • $\begingroup$ Sorry I was editing my previous reply when you wrote this. Please refer to my reply above. Thank you. $\endgroup$ – Deisler Apr 12 '18 at 20:58
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    $\begingroup$ Ok. Let’s just assume the two random variables, i.e. the length of rod and the coordinate of its near end, have their own probability density functions. Please ignore the ‘Gaussian’ part. $\endgroup$ – Deisler Apr 12 '18 at 21:54
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The terminology of probability--outcomes, events, and random variables--was invented to help with such problems. But let's begin with a picture of the situation:

Figure

This figure plots possible values of the endpoints $(X,Y).$ They are random variables. The axes are copies of the line; its origin is at the bottom left where the axes meet.

Pick a point $x$ on the line, as shown simultaneously on both axes.

  • Question 1 concerns the event $$\mathcal{E}_1:X \ge x$$ where the point $x$ is covered by the rod of length $X$.

  • Question 2 concerns the event $$\mathcal{E}_2:X \le x \le Y \text{ or } Y \le x \le X$$ where the point $x$ is covered by a rod with termini at $X$ and $Y$. This event is shown by the gray portion of the figure (which extends, in principle, infinitely far to the right and infinitely far upwards).

The distribution of $X$ is often described by stating what probability is assigned to the event $X\le x.$ This is the distribution function (aka CDF) of $X,$ $$F_X(x)=\Pr(X\le x).$$ In these terms, question (1) is answered using basic laws of probability (namely, the probabilities of disjoint events add) as

$$\eqalign{ \Pr(\mathcal{E}_1) &= \Pr(X=x\text{ or } X \gt x) \\ &= \Pr(X=x) + 1 - \Pr(X\le x) \\ &= \Pr(X=x) + 1 - F_X(x). }$$

The joint distribution of $(X,Y)$ can be described by stating what probability it assigns to infinite quadrants whose upper right corners are at a location $(x,y).$ This is the distribution function $$F_{XY}(x,y) = \Pr(X\le x\text{ and } Y \le y).$$

The figure shows we can express the probability of the gray region as equal to

  1. The probability of the entire region to the left of or at $X=x,$ equal to $F_X(x),$

  2. plus the probability of the entire region below or at $Y=x,$ equal to $F_Y(x),$

  3. minus the probability of the intersection of those regions (shown as the square with $(x,x)$ at its upper corner), equal to $F_{XY}(x,x).$

We see that the "interplay of events" is visually captured by this two-dimensional picture. This is a general, powerful technique for reasoning about two random variables together.

Thus, the general answer to the second question is

$$\Pr(\mathcal{E}_2) = F_X(x) + F_Y(x) - F_{XY}(x,x).$$

The termini are said to be independent when $F_{XY}(x,y) = F_X(x)F_Y(y).$ In that case the general answer can be reduced to a formula in terms of $F_X$ and $F_Y$ alone by substitution in its last term.

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  • $\begingroup$ Thank you so much for your explanations. I understand your answer to Question 1, however I do not quite get your answer to Question 2. If X represents the length of the rod, and Y the length of the 'void' between the origin and the near end of the rod, should the event in question 2 become Y<=x<=X+Y instead? $\endgroup$ – Deisler Apr 12 '18 at 23:27
  • $\begingroup$ Sorry I misread the definitions of your X and Y variables. If your X represents near end and Y the far end. Then is the probability function of Y the convolution of the rod length and the length of the ‘void’ between origin and the rod? So basically I just need to do the convolution step before using following your answers? $\endgroup$ – Deisler Apr 13 '18 at 0:02
  • $\begingroup$ The convolution won't work, because if you assume $X$ is near and $Y$ is far, then $X$ and $Y$ are (strongly) dependent, whence you must use the general form of the solution I originally derived. $\endgroup$ – whuber Apr 13 '18 at 14:38

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