# Let V denote rainfall volume and W denote runoff volume (both in mm). According to the article “Runoff Quality Analysis of Urban Catchments with Analytical Probability Models” (J. of Water Resource Planning and Management, 2006: 4–14), the runoff volume will be 0 if $V\ \leq\ v_d\$ and will $k\ (V\ -\ v_d) if\ V\ >\ v_d.\ Here\ v_d$ is the volume of depression storage (a constant) and k (also a constant) is the runoff coefficient. The cited article proposes an exponential distribution with parameter $\lambda\ for\ V.$ a. Obtain an expression for the cdf of W. [Note: W is neither purely continuous nor purely discrete, instead it has a “mixed” distribution with a discrete component at 0 and is continuous for values w > 0.] b. What is the pdf of W for w>0? Use this to obtain an exp

Question
Exponential models
Let V denote rainfall volume and W denote runoff volume (both in mm). According to the article “Runoff Quality Analysis of Urban Catchments with Analytical Probability Models” (J. of Water Resource Planning and Management, 2006: 4–14), the runoff volume will be 0 if $$\displaystyle{\left[{V}\ \leq\ {v}_{{d}}\ \right]}$$ and will $$\displaystyle{\left[{k}\ {\left({V}\ -\ {v}_{{d}}\right)}{\quad\text{if}\quad}\ {V}\ {>}\ {v}_{{d}}.\ {H}{e}{r}{e}\ {v}_{{d}}\right]}$$ is the volume of depression storage (a constant) and k (also a constant) is the runoff coefficient. The cited article proposes an exponential distribution with parameter $$\displaystyle{\left[\lambda\ {f}{\quad\text{or}\quad}\ {V}.\right]}$$
a. Obtain an expression for the cdf of W.
[Note: W is neither purely continuous nor purely discrete, instead it has a “mixed” distribution with a discrete component at 0 and is continuous for values $$\displaystyle{w}{>}{0}$$.]
b. What is the pdf of W for $$\displaystyle{w}{>}{0}$$? Use this to obtain an expression for the expected value of runoff volume.

2021-02-14

a) It is given that random variable V denotes the rainfall volume and has exponential distribution with parameter $$\displaystyle{\left[\lambda\right]}$$, then using the standard results derived in the book, it's pdf is given as:
$$f_\upsilon(\upsilon)\ =\ \begin{cases}0 & \upsilon < 0\\\lambda\ \cdot\ -e^{-\lambda \upsilon} & \upsilon \geq 0\end{cases}$$
And it's cdf is given as:
$$F_\upsilon(\upsilon)\ =\ \begin{cases}0 & \upsilon < 0\\1\ -e^{-\lambda \upsilon} & \upsilon \geq 0\end{cases}$$
It is also given that rv W denotes runoff volume and it's value depends on V as follows:
$$W\ =\ \begin{cases}0 & V \leq\ \upsilon_d\\k(V\ -\ \upsilon_d) & V\ > \upsilon_d\end{cases}$$
Let us denote cdf of W as $$\displaystyle{\left[{F}_{{w}}{\left({w}\right)}.\right]}$$
As we can see that W can never be less tan zero, hence
$$\displaystyle{\left[{P}{\left({W}\ {<}\ {0}\right)}\ =\ {0}\right]}$$
and using the definition of cdf, we can say that:
$$\displaystyle{\left[{F}_{{w}}{\left({w}\right)}\ =\ {0},\ {f}{\quad\text{or}\quad}\ {w}\ {<}\ {0}\right]}$$
As given in the note, $$\displaystyle{\left[{F}_{{w}}{\left({0}\right)}\ \ne{q}\ {0}\right]}$$
Which means $$\displaystyle{\left[{F}_{{w}}{\left({w}\right)}\right]}$$ is not continuous for all values of w. Hence
$$\displaystyle{\left[{F}_{{w}}{\left({0}\right)}\ {P}{\left({W}\right)}\ \leq\ {0}\right.}$$
$$\displaystyle=\ {P}{\left({W}\ =\ {0}\right)}\ +\ {P}{\left({W}\ {<}\ {0}\right)}$$
$$\displaystyle=\ {P}{\left({W}\ =\ {0}\right)}$$
$$\displaystyle=\ {P}{\left({V}\ \leq\ \upsilon_{{d}}\right)}$$
$$\displaystyle=\ {F}_{\upsilon}{\left(\upsilon_{{d}}\right)}$$
$$\displaystyle=\ {1}\ -\ {e}^{{-\lambda\upsilon_{{d}}}}{]}$$
Similiarly for $$\displaystyle{\left[{w}\ {>}\ {0}\right.}$$
$$\displaystyle{F}_{{w}}{\left({w}\right)}\ =\ {P}{\left({W}\ \leq\ {w}\right)}$$
$$\displaystyle=\ {P}{\left({W}\ {<}\ {0}\right)}\ +\ {P}{\left({w}\ =\ {0}\right)}\ +\ {P}{\left({0}\ \leq\ {W}\ {<}\ {w}\right)}$$
$$\displaystyle=\ {0}\ +\ {\left[{1}\ -\ {e}^{{-\lambda\upsilon_{{d}}}}\right]}\ +\ {P}\ {\left({0}\ \leq\ {k}{\left({V}\ -\ \upsilon_{{d}}\right)}\ \leq{w}\right)}$$
$$\displaystyle=\ {\left[{1}\ -\ {e}^{{-\lambda\upsilon_{{d}}}}\right]}\ +\ {P}\ {\left(\upsilon_{{d}}\ \leq\ {V}\ \leq\ \upsilon_{{d}}\ +\ {\frac{{{w}}}{{{k}}}}\right)}$$
$$\displaystyle=\ {\left[{1}\ -\ {e}^{{-\lambda\upsilon_{{d}}}}\right]}\ +\ {\left[{F}_{\upsilon}\ {\left(\upsilon_{{d}}\ +\ {\frac{{{w}}}{{{k}}}}\right)}\ -\ {F}_{\upsilon}{\left(\upsilon_{{d}}\right)}\right]}$$
$$\displaystyle=\ {\left[{1}\ -\ {e}^{{-\lambda\upsilon_{{d}}}}\right]}\ +\ {\left[{\left({1}\ -\ {\exp{{\left(-\lambda\upsilon_{{d}}\ +\ {\frac{{-\lambda{w}}}{{{k}}}}\right)}}}\right)}\ -\ {\left({1}\ -\ {e}^{{-\lambda\upsilon_{{d}}}}\right)}\right]}$$
$$\displaystyle=\ {1}\ -\ {\exp{\ }}{\left(-\lambda\upsilon_{{d}}\ +\ {\frac{{-\lambda{w}}}{{{k}}}}\right)}{]}$$
Finally $$\displaystyle{\left[{F}_{{w}}{\left({w}\right)}\right]}$$ can be written as:
$$F_w(w)\ =\ \begin{cases}0 & w < 0\\1\ -\ exp({-\lambda \upsilon_d + \frac{-\lambda w}{k}} & w \geq 0\end{cases}$$
Proposition: Let Z be a continuous rv with cdf $$\displaystyle{\left[\phi{\left({z}\right)}.\right]}$$ Then for any number a,
$$\displaystyle{\left[{P}{\left({Z}\ {>}\ \alpha\right)}\ =\ {1}\ -\ \phi{\left(\alpha\right)}\right.}$$
$$\displaystyle{P}{\left({Z}\ {<}\ \alpha\right)}\ =\ \phi{\left(\alpha\right)}{]}$$
(b)
To derive pdf of W from it's cdf, we recall following proposition:
Proposition: If X is a continuous rv with pdf f(x) and cdf F(x), then at every x at which the derative F'(x) exist,
$$\displaystyle{\left[{f{{\left({w}\right)}}}{\left({w}\right)}\ =\ {\frac{{{d}}}{{{d}{w}}}}\ {\left({1}\ -\ {\exp{{\left(-\lambda\upsilon_{{d}}+{\frac{{-\lambda{w}}}{{{k}}}}\right)}}}\right)}\right.}$$
$$\displaystyle={\frac{{{d}}}{{{d}{w}}}}\ {\left({1}\ -\ {e}^{{-\lambda\upsilon_{{d}}}}\cdot{\left({\frac{{-\lambda\omega}}{{{k}}}}\right)}\right)}$$
$$\displaystyle{{f}_{{w}}{\left({w}\right)}}\ =\ {\frac{{\lambda}}{{{k}}}}\ \cdot\ {e}^{{-\lambda\upsilon_{{d}}}}\ \cdot\ {\exp{\ }}{\left({\frac{{-\lambda{w}}}{{{k}}}}\right)}{]}$$
Overall pdf can be written as:
$$f_{w}(w)=\begin{cases}0 & w<0\\1-e^{-\lambda v_{d}} & w=0 \\ \frac{\lambda}{k}\cdot e^{-\lambda v_{d}}\cdot exp(\frac{-\lambda w}{k}) & w>0\end{cases}$$
Now we recall the definition of expexted value of a pdf:
Definition: The expected or mean value of a continuous rv X with pdf $${f{{\left({x}\right)}}}\ {i}{s}$$
$$\displaystyle{E}{\left({X}\right)}\ =\ {\int_{{-\infty}}^{{\infty}}}{x}\ \cdot\ {f{{\left({x}\right)}}}\ \cdot\ {\left.{d}{x}\right.}$$
Using this, we can write $$\displaystyle{\left[{E}{\left({W}\right)}\right]}$$ as:
$$\displaystyle{\left[{E}{\left({X}\right)}\ =\ {\int_{{-\infty}}^{{\infty}}}\omega\ \cdot\ {{f}_{{\omega}}{\left(\omega\right)}}\ \cdot\ {d}\omega\right.}$$
$$\displaystyle=\ {\int_{{{0}}}^{{\infty}}}\ {w}\ \cdot\ {\frac{{\lambda}}{{{k}}}}\ \cdot\ {e}^{{-\lambda\upsilon_{{d}}}}\ \cdot\ {\exp{{\left({\frac{{-\lambda{w}}}{{{k}}}}\right)}}}{d}{w}$$
$$\displaystyle=\ {\frac{{\lambda\ \cdot\ {e}^{{-\lambda\upsilon_{{d}}}}}}{{{k}}}}\ {\int_{{{0}}}^{{\infty}}}\ {w}\ \cdot\ {\exp{\ }}{\left({\frac{{-\lambda\omega}}{{{k}}}}{d}{w}\right)}{]}$$
Here we will use integration by parts, which can be summarized as follows:
Integration by parts: If u and v are function of x, then
$$\int_{a}^{b}\ u \upsilon^{\prime} \cdot\ dx\ =\ [u \upsilon]_{a}^{b}\ -\ \int_{a}^{b}\ u^{\prime} \upsilon\ \cdot\ dx$$
In our case, we take $$\displaystyle{\left[{u}\ =\ {w}\right]}{\quad\text{and}\quad}{\left[\upsilon\ =\ {\exp{\ }}{\left({\frac{{-\lambda{w}}}{{{k}}}}\right)}\right]}$$
Then the integral becomes:
$$\displaystyle{\left[{E}{\left({W}\right)}\ =\ {\frac{{\lambda\ \cdot\ {e}^{{-\lambda\upsilon_{{d}}}}}}{{{k}}}}\ {\left({\left[-{w}\ \cdot\ {\frac{{{k}}}{{\lambda}}}\ \cdot\ {\exp{\ }}{{\left({\frac{{-\lambda{w}}}{{{k}}}}\right]}_{{{0}}}^{{\infty}}}\ +\ {\frac{{{k}}}{{\lambda}}}\ \cdot\ {\int_{{{0}}}^{{\infty}}}\ {\exp{{\left({\frac{{-\lambda{w}}}{{{k}}}}\right)}}}{d}{w}\right)}\right.}\right.}$$
$$\displaystyle{E}{\left({W}\right)}\ =\ {\frac{{\lambda\ \cdot\ {e}^{{-\lambda\upsilon_{{d}}}}}}{{{k}}}}\ {\left({{\left[-{w}\ \cdot\ {\frac{{{k}}}{{\lambda}}}\ \cdot\ {\exp{\ }}{\left({\frac{{-\lambda{w}}}{{{k}}}}\right)}\right]}_{{{0}}}^{{\infty}}}\ -\ {\frac{{{k}^{{2}}}}{{\lambda^{{2}}}}}\ \cdot\ {{\left[{\exp{{\left({\frac{{-\lambda{w}}}{{{k}}}}\right)}}}\right]}_{{{0}}}^{{\infty}}}\right)}$$
$$\displaystyle{E}{\left({W}\right)}\ =\ {\frac{{\lambda\ \cdot\ {e}^{{-\lambda\upsilon_{{d}}}}}}{{{k}}}}\ {\left({\left[{0}\ -\ {0}\right]}\ -{\frac{{{k}^{{2}}}}{{\lambda^{{2}}}}}\ \cdot\ {\left[{0}\ -\ {1}\right]}\right)}$$
$$\displaystyle{E}{\left({W}\right)}\ =\ {\frac{{\lambda\ \cdot\ {e}^{{-\lambda\upsilon_{{d}}}}}}{{{k}}}}\ {\left({\frac{{{k}^{{2}}}}{{\lambda^{{2}}}}}\right)}$$
$$\displaystyle{E}{\left({W}\right)}\ =\ {\frac{{\lambda\ \cdot\ {e}^{{-\lambda\upsilon_{{d}}}}}}{{\lambda}}}$$

### Relevant Questions

As depicted in the applet, Albertine finds herself in a very odd contraption. She sits in a reclining chair, in front of a large, compressed spring. The spring is compressed 5.00 m from its equilibrium position, and a glass sits 19.8m from her outstretched foot.
a)Assuming that Albertine's mass is 60.0kg , what is $$\displaystyle\mu_{{k}}$$, the coefficient of kinetic friction between the chair and the waxed floor? Use $$\displaystyle{g}={9.80}\frac{{m}}{{s}^{{2}}}$$ for the magnitude of the acceleration due to gravity. Assume that the value of k found in Part A has three significant figures. Note that if you did not assume that k has three significant figures, it would be impossible to get three significant figures for $$\displaystyle\mu_{{k}}$$, since the length scale along the bottom of the applet does not allow you to measure distances to that accuracy with different values of k.
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(a) What is thedensity for X? Sketch a graph of the density function. Indicate onthis graph the probability that X lies between 1.1 and 1.9. Findthis probability.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
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(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
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$$\displaystyle{F}_{{{d}{r}{u}{g}}}={C}_{{d}}{A}{v}^{{2}}$$
where A is the cross-sectional area of the vehicle and $$\displaystyle{C}_{{d}}$$ is called the coefficient of drag.
Part A:
Consider a vehicle moving with constant velocity $$\displaystyle\vec{{{v}}}$$. Find the power dissipated by form drag.
Express your answer in terms of $$\displaystyle{C}_{{d}},{A},$$ and speed v.
Part B:
A certain car has an engine that provides a maximum power $$\displaystyle{P}_{{0}}$$. Suppose that the maximum speed of thee car, $$\displaystyle{v}_{{0}}$$, is limited by a drag force proportional to the square of the speed (as in the previous part). The car engine is now modified, so that the new power $$\displaystyle{P}_{{1}}$$ is 10 percent greater than the original power ($$\displaystyle{P}_{{1}}={110}\%{P}_{{0}}$$).
Assume the following:
The top speed is limited by air drag.
The magnitude of the force of air drag at these speeds is proportional to the square of the speed.
By what percentage, $$\displaystyle{\frac{{{v}_{{1}}-{v}_{{0}}}}{{{v}_{{0}}}}}$$, is the top speed of the car increased?
Express the percent increase in top speed numerically to two significant figures.
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b. Describe what new forwarding paths are needed for the rearranged pipeline by stating the source, destination, and information transferred on each needed new path.
c. For the reordered stages of the RISC pipeline, what new data hazards are created by this addressing mode? Give an instruction sequence illustrating each new hazard.
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A. Find the radii of the two "daughter" nuclei of charge+46e.
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This problem is about the equation
dP/dt = kP-H , P(0) = Po,
where k > 0 and H > 0 are constants.
If H = 0, you have dP/dt = kP , which models expontialgrowth. Think of H as a harvesting term, tending to reducethe rate of growth; then there ought to be a value of H big enoughto prevent exponential growth.
Problem: find acondition on H, involving $$\displaystyle{P}_{{0}}$$ and k, that will prevent solutions from growing exponentially.

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f(x)-?
g(x)-??
$$\begin{array}{|l|l|l|}\hline X&-2&-1&0&1&2\\\hline f(x)&1.125&2.25&4.5&9&18\\\hline g(x)&16&8&4&2&1\\\hline\end{array}$$

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