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Let V denote rainfall volume and W denote runoff volume (both in mm). According to the article 鈥淩unoff Quality Analysis of Urban Catchments with Analytical Probability Models鈥� (J. of Water Resource Planning and Management, 2006:4-14), the runoff volume will be 0 if V\拢{\nu _d}and will be \kappa (V - {\nu _d})if{\text{ }}V > {\nu _d} . Here {\nu _d} is the volume of depression storage (a constant), and (also a constant) is the runoff coefficient. The cited article proposes an exponential distribution with parameter \lambda {\text{ }}for{\text{ }}V .

a. Obtain an expression for the cdf of W. [Note: Wis neither purely continuous nor purely discrete; instead, it has a 鈥渕ixed鈥� distribution with a discrete component at 0 and is continuous for valuesw{\text{ }} > {\text{ }}0.]

b. What is the pdf of W for w{\text{ }} > {\text{ }}0? Use this to obtain an expression for the expected value of runoff volume.

Step by step solution

01

Definition of probability

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Determining the cdf of W                

(a) Given that random variable V signifies rainfall volume and has an exponential distribution with parameter \lambda , the pdf is as follows:

{f_v}(v) = \left\{ {\begin{array}{{}{}}{v < 0} \\{\lambda \times {e^{ - \lambda v}}v0}

\end{array}} \right.

Its cdf is as follows:

{F_v}(v) = \left\{ {\begin{array}{{}}{v < 0} \\{1 - {e^{ - \lambda v}}v0}\end{array}} \right.

It's also worth noting that rv's stands for runoff volume, and its value is determined by V as follows:

W = \left\{ {\begin{array}{{}}{V\拢{v_d}} \\{k\left( {V - {v_d}} \right)V > {v_d}}

\end{array}} \right.

Let's call the of

As can be seen, can never be less than zero, and so

We may state this using the definition of :

, as stated in the note.

This suggests that is not continuous for all values. Hence

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