Question

Let R be a uniform random variable on [0, 1] and let g(x) = 1 − x. Find the probability
density function of the random variable S = g(R) using the density function method.

Answer 1

GHI!!

Answer 2

you got a worked out example? i think the first step it to look at \(g(x)=1-x\) and find \(g^{-1}\)

Answer 3

which in this case is just \(g\) again, because it is its own inverse

Answer 4

then find \(|\frac{dg^{-1}}{dx}|\) which is just \(-1\) here
does this ring a bell?

Answer 5

no i dont know how to go about with this question ahha

Answer 6

isnt there a formula where you do the inverse X derivative of the inverse?

Answer 7

I'm not sure what you mean by "density function method", but it could just be the name you learned for the method I show here (which my reference calls the method of distributions, while Misty's comments allude to what it calls the method of transformations).

Since \(R\) is uniformly distributed, its PDF is
\[f_R(r)=\begin{cases}1&\text{for }0<r<1\\[1ex]0&\text{otherwise}\end{cases}\]Then the CDF for \(R\) is given by
\[F_R(r)=\mathbb P(R\le r)=\begin{cases}\displaystyle \int_{-\infty}^r f_R(t)\,\mathrm dt=\color{red}0&\text{for }r\le0\\[1ex]
\displaystyle\int_{-\infty}^rf_R(t)\,\mathrm dt=\int_0^r\mathrm dt=\color{red}r&\text{for }0<r<1\\[1ex]
\displaystyle\int_{-\infty}^rf_R(t)\,\mathrm dt=\color{red}1&\text{for }r\ge1\end{cases}\]
Now, since \(S=g(R)=1-R\), the CDF for \(S\) is
\[\begin{align*}
F_S(s)&=\mathbb P(S\le s)\\[1ex]
&=\mathbb P(1-R\le s)\\[1ex]
&=\mathbb P(R\ge1-s)\\[1ex]
&=1-\mathbb P(R\le1-s)\\[1ex]
&=1-F_R(1-s)
\end{align*}\]Knowing the definition of \(F_R(r)\), you have
\[F_R(1-s)=\begin{cases}0&\text{for }1-s\le0\\[1ex]1-s&\text{for }0<1-s<1\\[1ex]1&\text{for }1-s\ge1\end{cases}\]There should be enough here to find out what \(F_S(s)\) is.

Answer 8

Well, that actually wouldn't finish the problem, since you need to find \(f_S(s)\), but that's just a matter of finding the derivative \(\dfrac{\mathrm d}{\mathrm ds}F_S(s)\).

Answer 9

okay. Thanks. Let met try it out

Answer 10

what you're using is called the distribution function method. My teacher wants us to use the density function method (transformation method). So basically I need to solve the problem using Misty's method (transformation method)

Answer 11

Okay, in case earlier comments weren't clear, I'll be happy to outline the method when I get a chance to refresh my memory on the method. But first I have some business to take care of at the moment.

Answer 12

okay

Answer 13

So the method of transformations offers you a direct way of determining \(f_S(s)\) given \(f_R(r)\) and \(S=g(R)\) provided that \(g\) is invertible, which as Misty said, it is.
\[g(R)=S=1-R\implies g^{-1}(S)=R=1-S\]The PDF of \(S\) is then
\[f_S(s)=f_R\left(g^{-1}(s)\right)\left|\frac{\mathrm dg^{-1}(s)}{\mathrm ds}\right|=f_R(1-s)\]since the derivative of the inverse function is \(-1\).

So if
\[f_R(r)=\begin{cases}1&\text{for }0<r<1\\[1ex]0&\text{otherwise}\end{cases}\]then what's \(f_S(s)=f_R(1-s)\)?

Answer 14

i got to the formula but i dont understand what to do after that. How did you get the following
fR(r)={1 for 0<r<1
0 otherwise

Answer 15

That's the definition of the (standard) uniform distribution, which is how \(R\) is distributed.

Answer 16

Given this definition, you know that
\[f_S(s)=f_R(1-s)=\begin{cases}1&\text{for }0<1-s<1\\[1ex]0&\text{otherwise}\end{cases}=\begin{cases}1&\text{for }0<s<1\\[1ex]0&\text{otherwise}\end{cases}\]so in fact \(S\) has the same distribution as \(R\).

Just to verify the result, let's compute the derivative of the CDF we found with the other method:
\[\begin{align*}
f_S(s)&=\frac{\mathrm d}{\mathrm ds}F_S(s)\\[1ex]
&=\frac{\mathrm d}{\mathrm ds}\left[1-F_R(1-s)\right]\\[1ex]
&=-\frac{\mathrm d}{\mathrm ds}F_R(1-s)\\[1ex]
&=\begin{cases}0&\text{for }1-s<0\iff \text{for }s>1\\[1ex]
-\dfrac{\mathrm d}{\mathrm ds}[-s]&\text{for }0<1-s<1\iff\text{for }0<s<1\\[1ex]
0&\text{for }1-s>1\iff\text{for }s<0
\end{cases}\\[1ex]
&=\begin{cases}1&\text{for }0<s<1\\[1ex]0&\text{otherwise}\end{cases}
\end{align*}\]and indeed, \(S\) and \(R\) have the same distribution.