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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, $4-14$ , the runoff volume will be 0 if $V \leq v_{d}$ and will be $k\left(V-v_{d}\right)$ 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 $W .[N o t e : W$ 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 expression for the expected value of runoff volume.

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

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Missouri State University

University of North Carolina at Chapel Hill

Cairn University

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So in this particular problem, we've been given the conditions for the c d f of w. So for the part a we know that w is a known 0 value. So, therefore, i can see that w is greater than or equal to 0 and it is not 0 for this part. So, therefore, for w less than 0, the c r f of w, we can write it as b of 4 capital w less than equal to the small value w, which can also be written as a c d f of f of w. And that part we need to simplify for different values now, so first i want to find probability for w is equal to exact 0 value is the first part we are going to find so to find out. A probability of w is equal to 0. The condition given in the question is that it is v less than is equal to v d, which means that this would be integral from 0 to v d and for the expansion distribution. The value of f of x is lambda into e raised to minus lambda x dx. So if you apply the integration, this is lambda e raised to minus lambda x, upon minus lambda limit is from 0 to e d, so lambda is canceled and if i substitute the limits, i get 1 minus e raise to minus lambda v d. So this is the probability when w is equal to 0. Now next is i'm supposed to find the probability where the capital w will lie between the value 0 and small w okay. So, to start with, as for the conditions that are given in the question, this we can write is equal to the probability of 0 is less than capital. W, for this condition is k into v minus v d and less than the small value. So now we can rewrite this part like this, where we can multiply k inside b d and then divide for by so the expression i'm going to do that we have 0 less than we have k v minus k. V d is less than w. Okay. So then, further, if i divide that by k, what we will have here is 0 less than v. Minus v d is less than w by k. Okay and further, i'm going to add trod with v d. So we have probability of v. D is less than b is less than this is w, plus we have, i mean this will be w by k plus dkyso now, based on this condition, we have to integrate from v d to this is w plus k, v d upon k for the whole Lambda e raise to minus lambda x dx. So if i continue with the integration to keep lambda as it is, integration is raised to minus lambda x, upon minus lambda. Limits are from v d to this is w plus k, v d upon k, so lambda gets canceled here and the evaluated expression that we will get now is e raised to minus k v d- i'm sorry! This is e. Raised 2 minus not k, 3 lambda, v d and minus, we will have e raised to minus w plus k v d by k, so this was for the condition between 0 and thema. W now we're talking about the values for w is greater than or equal to 0. Then, in this case, the c d f is defined as again, f of w will be p of w less than equal to w. So this part i can take this as p of w is 0 plus the value of p between 0 and small w. So we found both the values if i substitute here- 1 minus e raised to minus lambda d plus. This is e raised to minus lambda d: minus e, raise to minus w plus k, v d upon k, so you can see it. We can this part. So what remains? Is 1 minus e raised to minus w plus k, v d upon k so based on this, i can therefore conclude the serf, as f of w can be written here as 0. For c w is less than 0, and this is 1 minus e raised to minus w plus k v d by k 4 w greater than equal to 0 point now going on to the part b. They also tried to find the p d f for this particular part, so the p d f, o w for w greater than 0, will be f, o w. We can take this as derivative of the c d f. So if i take the derivative of this expression with this is 1 minus e raise to minus lambda, we have w plus k v d upon kohei differentiate. This will be minus of e, raise to minus lambda into w plus k v d by k and into the inner derivative with w will be minus lambda by k. So the final value for this i'm getting here is lambda by k into e raised to minus lambda, w plus k v d by k. So this is the p d f for the given particular part. Now we also know that w is a mixed distribution, so we can say that, since w is a mixed distribution as given the question, it's a combination of discrete and continuous, both so now to find out the expectation for known to use the concept of discrete as Well, as continues so, then we were going to write expectation of so for discrete. I'M going to take this as x into p of xi, be 0 into p of w is equal to 0, and for this part this will be continuous. So we have integral 0 to infinity, will be w into f of woitin into f f w we've just found out, so that's actually lambda by k and to we have e raised to minus lambda, and this is w by k plus v d with d w. So i'm going to take lambda by k e raised to give minus lambda in the bracket. We have minus lambda d, constant, so taking that part outside here and integral is for 0 to infinity between w and e raised to minus lambda w by k d w. So applying the integration by parts now over here we're going to get into integration, for this is e raised to minus lambda w by k upon minus lambda by k limit from 0 to infinity, integral 0 to infinity derivative of w will be 1 and again. Integration of e raise to is divided by minus lambda w upon ki. Sorry remove this w here. We'Re integrating with that. So lambda by k is a constant and with d w p. So further, if we take minus lambda by k common throughout, so we'll have ink by lambda outside which will get canceled, and we will have e raised to minus lambda d as it is and in the bracket here the upper limit. This will change to 0 because he raised to minus infinity is 0 and was here is 0. The first term is 0 directly. This will change to plus we have k upon lambda, so i'm keeping lambda by k here as of now because canceled anyway, and for this we're again doing integration. So you will have e raised to minus lambda w by k upon minus lambda by k from 0 to infinity. So i can cancel this here and now. After putting the limits, we have lambda by k, e raise to minus lambda v d and the upper limit e raised to minus infinity is 0 and minus e raised to 0 is 1, and this divided by minus 1, so minus 1 gets canceled and orther is A correction here: this doesn't cancel it. In fact, they get multiplied. So his lambda, by k into i, have k square by lambda square, and now, when you put the limits, you'll have 1 minus 0. So, if you'll now simplify this part, the final answer for the expectation of that i'm getting is k by lambda e raised to minus lambda v d. So this is the final answer for expectation of.

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