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This is a question from an A level textbook on continuous random variables. It states that the CRV $T$ has pdf $f(t)=0.5 ~for~ 1<t<3$ and then goes on to ask us to find the CDF (easy peasy $F(t)=\frac{1}{2} (t-1)$ and to show that the probability of selecting two independent observation less than 2.5 is $\frac{9}{16}$ (again, fine, $F(2.5)\times F(2.5)$). The third part, however, is a bit weird. I'll quote exactly

"$S$ is the larger of two independent observations of $T$. By considering the CDF of $S$ show that $S$ has pdf $g(s)=\frac{s-1}{2}$ and then the details about the ranges it applied to"

I couldn't get anywhere with this and cheated looking up their worked solutions which amounted to

$P(S=s)=P(T=s)\times P(T\leq S)\times 2=0.5\times \frac{s-1}{2}\times 2=\frac{s-1}{2}$

In other words, you pick a value of $T=s$ and the next one needs to be smaller than it ($T\leq s$) but it could be the other way around (hence $\times 2$).

Now, I'm really concerned about $P(T=s)$, surely this is zero?

Your help will be much appreciated.

Jonathan Andrews
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2 Answers2

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Another approach.

If both observations $s_1$ and $s_2$ are smaller or equal than s, then $\max(s_1,s_2)\leq s$.

Thus you obtain the cdf by calculating $[P(T\leq s)]^2$. Let $G(s)$ denote the corresponding cdf, then

$$G(s)=\left(\frac{1}{2} (s-1)\right)^2 \mathbb 1_{\{1\leq s\leq 3\}} $$

To obtain the pdf you differentiate the cdf w.r.t. $s$

Henry
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callculus42
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You raise an excellent point. Their derivation here is fine, but looks unrigorous unless interpreted correctly.

This procedure is analogous to a calculation of the form “$\mathrm{d}u=5u\,\mathrm{d}x$” when doing a substitution in an integral. Strictly speaking, both the LHS and RHS don’t make sense unless you interpret them in a certain way. Here, I’m interpreting $P(S=s)$ as $g(s)=\mathrm{d}F_S$ where $F$ is the cumulative distribution function (this is all similar to Riemann-Stieltjes integration and some ideas from measure theory, eg the Radon-Nikodym derivative: beyond A level, but interesting further reading, perhaps...). Their use of “$P(T=s)$” rather than: “$\operatorname{pdf}_T(s)$” is possibly adding to the confusion, since strictly the probability is zero (as you say) while the probability density (the “$5u$” in: $5u\,\mathrm{d}x$) is not zero.

It might make more sense if you determine the cumulative distribution function for $S$ first. Unfortunately, to do so, we’ll still need to mess round with the same “zero” expression, but view it as moving from the discrete case (lots of sums) to a continuous case. It’s better to use density function notation though, since that’s what we are integrating (the distinction between the probability and the density is very important). $$\begin{align}P(S\le s)&=\int_1^s2\operatorname{pdf}_T(s’)\operatorname{cdf}_T(s’)\,\mathrm{d}s’\\&=\int_1^s2(1/2)(1/2)(s’-1)\,\mathrm{d}s’\\&=\frac{1}{4}(s-1)^2\end{align}$$

Now differentiate this expression to find $g(s)$.

FShrike
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    +1 I was going to comment on the final equation but you've already fixed it. – Suzu Hirose Aug 18 '22 at 06:59
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    Thank you, I'll work on this. My immediate reaction is, although, my student very smart, they are not university students 9they are 18 years old) and this distinction is subtle. Once upon a time, I cover CRV starting with CDF (makes more sense to me this way) and eventually leading to pdfs but the texts I use don't really facilitate that – Jonathan Andrews Aug 18 '22 at 07:10
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    @JonathanAndrews In which case, I recommend appealing to the density of $S,T$. Don’t write: “$P(S=s)=2P(T=s)P(T\le s)$” for your students, since really both sides are zero and the equation isn’t meaningful: write: “$\operatorname{pdf}_S(s)=2P(T\le s)\times\operatorname{pdf}_T(s)$”, which is better, as an equation of densities. – FShrike Aug 18 '22 at 07:20
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    Thank you, the difficulty is that I don't know the notation you've used and it's certainly not required in the curriculum I cover. We're pretty vague on what is meant by a pdf - I have to do more reading on it myself to be honest. In this context, it seems a terrible question - if I reproduced their answer, one of my students would certainly raise the question I asked and I couldn't answer it! I'm sure I can work on understanding pdfs better but, what for? My own education, fine but it's of little (no?) value to my students – Jonathan Andrews Aug 18 '22 at 07:25
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    @JonathanAndrews I don’t know your students or your course. However I am in the British system myself and I’m fairly sure (OCR MEI Further Maths Statistics Major course) that familiarity with pdfs and pdf/cdf notation is important. After all, pdfs are used in pretty much all questions concerning continuous random variables, and continuous random variables can only be dealt with through cdf, pdf (in this context of A-level maths). So, my notation is just: $\operatorname{pdf}_T(x)$ is the probability density function for the random variable $T$, evaluated at $x$. – FShrike Aug 18 '22 at 07:40
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    In this context, probability is to probability density what integration is to differentiation. Probability density tells you how to “weight” your “sum of chances”: begin with the discrete case, where one uses a probability “mass” function (same thing, really), then pass to an integral (A-level students are familiar with the concept but it doesn’t need to be rigorously explained, that waits until university) – FShrike Aug 18 '22 at 07:42
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    Of course, they do know the notation and thanks for the clarification. – Jonathan Andrews Aug 18 '22 at 07:44
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    @JonathanAndrews: There are two things I want to comment on: (1) There is no meaning to a "continuous random variable" without a notion of CDF or PDF, so being "vague on what is meant by a PDF" is really silly if one wants to do any mathematics with continuous random variables. A PDF for a random variable $v$ is simply a function such that its integral over a region gives the probability that $v$ has instantiated value in that region. (2) If you do not understand something well, you cannot teach it well. – user21820 Aug 18 '22 at 15:22