Question

I can't figure out how (c-a)/(b-a) + the integral of 2(b-x)/(b-a)(b-c) from c to x equals
1 - ((b-x)^2/(b-a)(b-c))

For context, I'm trying to see how the pdf (prob. density function) of the Triangular Distribution for c< x < b equals the CDF function given above.
https://en.wikipedia.org/wiki/Triangular_distribution

Answer 1

Typo, I meant that I'm trying to see how the integral of the 1-((b-x).... function equals the cdf for c < x< b

Answer 2

From the wiki, for \(c<x\le b\),
\[f(x)=\frac{2(b-x)}{(b-a)(b-c)}\]
By the definition of the CDF, you have \(F(x)=\displaystyle\int_{-\infty}^xf(t)\,dt\), i.e. for \(c<x\le b\) you have
\[F(x)=\color{red}{\int_{-\infty}^cf(t)\,dt}+\color{blue}{\int_c^x\frac{2(b-t)}{(b-a)(b-c)}\,dt}\]
Evaluating the first integral is simple enough:
\[f(x)=\begin{cases}
\dfrac{2(x-a)}{(b-a)(c-a)}&\text{for }a\le x<c\\\\
2(b-a)&\text{for }x=c\\\\
\dfrac{2(b-x)}{(b-a)(b-c)}&\text{for }c< x\le b\\\\
0&\text{elsewhere}
\end{cases}\]
which gives
\[0+\int_{a}^c\frac{2(x-a)}{(b-a)(c-a)}\,dx=\frac{1}{(b-a)(c-a)}\left[(x-a)^2\right]_a^c=\color{red}{\frac{c-a}{b-a}}\]
For the second integral, you have
\[\begin{align*}\int_c^x\frac{2(b-t)}{(b-a)(b-c)}\,dt&=\int_x^c\frac{2(t-b)}{(b-a)(b-c)}\,dt\\\\
&=\frac{1}{(b-a)(b-c)}\left[(t-b)^2\right]_x^c\\\\
&=\frac{(c-b)^2}{(b-a)(b-c)}-\frac{(x-b)^2}{(b-a)(b-c)}\\\\
&=\color{blue}{\frac{b-c}{b-a}-\frac{(x-b)^2}{(b-a)(b-c)}}
\end{align*}\]
Adding the first blue term to the red term, you have
\[\color{red}{\frac{c-a}{b-a}}+\color{blue}{\frac{b-c}{b-a}}=\frac{b-a}{b-a}=1\]
And so
\[F(x)=1-\frac{(x-b)^2}{(b-a)(b-c)}\]

Answer 3

Thanks for the prompt response, Siths!
Two questions:
1. Where did you get t from? (I assume it's a just a variable to differentiate from x, which we substitute in later).
2. Doesn't the integral of the numerator 2(t-b) = t^2 - 2bt ? Confused how you're able to keep b inside the parentheses when integrating?

Answer 4

Also, how does \[ \frac{ (c-b)^{2} }{ (b-a)(b-c) } = \frac{ b-c }{ b-a }\] ?

Answer 5

1. \(t\) is used as a dummy variable, or placeholder. At any given moment, we could certainly write
\[\int_c^xf(x)\,dx\]
but this can lead to some confusion. What does it mean to integrate with respect to one variable while also integrating over an interval containing that same variable? To get around this, we simply replace the integrating variable with another symbol, and I've chosen \(t\).


2. We're both right here. Notice:
\[\int (x-k)\,dx=\int x\,dx-k\int\,dx=\frac{1}{2}x^2-kx+C\]
while
\[\int(x-k)\,dx=\frac{1}{2}(x-k)^2+C=\frac{1}{2}x^2-kx+\frac{1}{2}k^2+C\]
The results differ by a constant \(\dfrac{1}{2}k^2\), which can be absorbed into the constant of integration \(C\), so in fact the results are the same.

The second integration approach can also be viewed with a substitution, say \(u=x-k\), such that \(du=dx\), and so
\[\int(x-k)\,dx=\int u\,du=\frac{1}{2}u^2+C=\frac{1}{2}(x-k)^2+C\]

3. This is a result of some simple algebra:
\[(c-b)^2=\bigg(-(b-c)\bigg)^2=(-1)^2(b-c)^2=(b-c)^2\]
and so
\[\frac{ (c-b)^{2} }{ (b-a)(b-c) } = \frac{ (b-c)^2 }{( b-a)(b-c) }=\frac{b-c}{b-a}\]

Answer 6

OK, got it. Thank you very much Siths for the thorough explanation!

Answer 7

You're welcome!