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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.

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2 Answers 2

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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. $s$

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    $\begingroup$ Thank you, this looks a more accessible method. $\endgroup$ Aug 18 at 7:29
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    $\begingroup$ Did you mean: $P(T\le s)^2$, and: $\max(s_1,s_2)\le s$? $\endgroup$
    – FShrike
    Aug 18 at 7:34
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    $\begingroup$ @FShrike I suspect the answer is "yes" so I made the edit $\endgroup$
    – Henry
    Aug 18 at 12:34
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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 $g(s)$.

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    $\begingroup$ +1 I was going to comment on the final equation but you've already fixed it. $\endgroup$ Aug 18 at 6:59
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    $\begingroup$ 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 $\endgroup$ Aug 18 at 7:10
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    $\begingroup$ @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. $\endgroup$
    – FShrike
    Aug 18 at 7:20
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    $\begingroup$ 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 $\endgroup$ Aug 18 at 7:25
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    $\begingroup$ @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 $x$. $\endgroup$
    – FShrike
    Aug 18 at 7:40

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