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Probability - Uniform distribution word problem. |
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| Aug5-12, 03:52 PM | #1 |
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Probability - Uniform distribution word problem.
1. The problem statement, all variables and given/known data
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? 2. Relevant equations 3. 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. [tex]T = \begin{cases} 30 - (x + 10) &\text{if } 0\leq x \leq 20 \\ 0 &\text{if } 20<x\leq 25 \end{cases}[/tex] [tex]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} [/tex] [tex] \mathbb{P}[20-x \leq t] = 1 - \mathbb{P}[x<20 -t][/tex] [tex]= 1 - \int_{0}^{20-t} \frac{1}{25} dx[/tex] [tex] = \frac{5+t}{25}[/tex] [tex]\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}[/tex] This would imply no density, but that seems plausible given that for [tex]T=0[/tex], [tex] 20<X\leq 25[/tex]. |
| Aug6-12, 11:07 AM | #2 |
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Recognitions:
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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 probabilty 20/25, T has a density on (0,20); that density would be [tex] g(t) = \frac{25}{20} \frac{d}{dt} \text{P}\{ T \leq t \}. [/tex] Note that the normalization gives [itex] \int_0^{20} g(t) \, dt = 1.[/itex] Sometimes Physicists and Engineers think of such random variables as having densities involving the Dirac delta-function: [tex] \text{density of }T = f(t) = (1/5)\delta(t) + (4/5) g(t). [/tex] 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: [tex] \text{cdf} = F(t) = (1/5)H(t) + (4/5) G(t),[/tex] where H is the Heaviside function H(w) = 0 for w < 0 and H(w) = 1 for w ≥ 0. RGV |
| Aug8-12, 04:31 PM | #3 |
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Thank you very much. I understand the concept of mixed distributions much better now.
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