Probability - Uniform distribution word problem.

[IniquiTrance]

Homework Statement

Jake leaves home at a random time between 7:30 and 7:55 a.m.
(assume the uniform distribution) and walks to his office. The walk takes 10
minutes. Let T be the amount of time spends in his office between 7:40 and
8:00 a.m.. Find the distribution function F_T of T and draw its graph. Does F_T
have a density?

Homework Equations

The Attempt at a Solution

This is what I've come up with so far:

Let X be the number of minutes past 7:30 that he leaves his house.

T = \begin{cases} 30 - (x + 10) &\text{if } 0\leq x \leq 20 \\ 0 &\text{if } 20<x\leq 25 \end{cases}

F_T(t) = \mathbb{P}[T\leq t] = \begin{cases} \mathbb{P}[20-x\leq t] &\text{if } 0\leq t \leq 20 \\ 0 &\text{if } t<0 \\ 1 &\text{if } t>20\end{cases}

\mathbb{P}[20-x \leq t] = 1 - \mathbb{P}[x<20 -t]
= 1 - \int_{0}^{20-t} \frac{1}{25} dx
= \frac{5+t}{25}

\therefore F_T(t) = \mathbb{P}[T\leq t] = \begin{cases} \frac{5+t}{25} &\text{if } 0\leq t \leq 20 \\ 0 &\text{if } t<0 \\ 1 &\text{if } t>20\end{cases}

This would imply no density, but that seems plausible given that for T=0, 20<X\leq 25.

 

[Ray Vickson]

Homework Statement

Jake leaves home at a random time between 7:30 and 7:55 a.m.
(assume the uniform distribution) and walks to his office. The walk takes 10
minutes. Let T be the amount of time spends in his office between 7:40 and
8:00 a.m.. Find the distribution function F_T of T and draw its graph. Does F_T
have a density?

Homework Equations

The Attempt at a Solution

This is what I've come up with so far:

Let X be the number of minutes past 7:30 that he leaves his house.

T = \begin{cases} 30 - (x + 10) &\text{if } 0\leq x \leq 20 \\ 0 &\text{if } 20<x\leq 25 \end{cases}

F_T(t) = \mathbb{P}[T\leq t] = \begin{cases} \mathbb{P}[20-x\leq t] &\text{if } 0\leq t \leq 20 \\ 0 &\text{if } t<0 \\ 1 &\text{if } t>20\end{cases}

\mathbb{P}[20-x \leq t] = 1 - \mathbb{P}[x<20 -t]
= 1 - \int_{0}^{20-t} \frac{1}{25} dx
= \frac{5+t}{25}

\therefore F_T(t) = \mathbb{P}[T\leq t] = \begin{cases} \frac{5+t}{25} &\text{if } 0\leq t \leq 20 \\ 0 &\text{if } t<0 \\ 1 &\text{if } t>20\end{cases}

This would imply no density, but that seems plausible given that for T=0, 20<X\leq 25.

You have not posted an actual question, and you seem to have done the problem satisfactorily. However, it would have been more straightforward to base the analysis on arrival time instead of departure: his arrival is uniform from 7:40 to 7:65 (if you will allow minutes > 60 for convenience), and you want to know the distribution of his office time duration between 7:40 and 7:60.

So, if 7:40 <--> 0, we have X~Unif(0,25) and you want the distribution of T = max(20-X,0). Of course, T is "mixed", with a finite probability mass at t = 0: P(T = 0} = 5/25 (5 minutes out of 25). For 0 < t < 20 you can think of T as having a density if you want, but that just means that the cdf P{T ≤ t} is differentiable in that interval. You could also think of T as being a probabilistic mixture of two distributions: with probability 5/25, T is derministic at 0 (that is, is a degenerate random variable concentrated at 0); with probability 20/25, T has a density on (0,20); that density would be g(t) = \frac{25}{20} \frac{d}{dt} \text{P}\{ T \leq t \}. Note that the normalization gives \int_0^{20} g(t) \, dt = 1. Sometimes Physicists and Engineers think of such random variables as having densities involving the Dirac delta-function: \text{density of }T = f(t) = (1/5)\delta(t) + (4/5) g(t).
However, not everybody would like or accept this "density", so be careful to gauge your audience before presenting it. However, it is always acceptable to view the cdf as a mixture:
\text{cdf} = F(t) = (1/5)H(t) + (4/5) G(t),
where H is the Heaviside function H(w) = 0 for w < 0 and H(w) = 1 for w ≥ 0.
RGV

 

[IniquiTrance]

Thank you very much. I understand the concept of mixed distributions much better now.