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(c) Let X,xz" Xkbeii.d. Exp(1) , and let Mk max( X, Xc) (Maximu o Xh_X)_ Fiud the prohability deusity function of Mk:(Hint: P(min(Xh,Xz,Xs) > k) = P(Xi >...

Question

(c) Let X,xz" Xkbeii.d. Exp(1) , and let Mk max( X, Xc) (Maximu o Xh_X)_ Fiud the prohability deusity function of Mk:(Hint: P(min(Xh,Xz,Xs) > k) = P(Xi > k,x >k,X > k), how ahout max ?)(d) Show that k-0,the CDF o Z = Mk ln(k) is the CDF of Y.Hint: Note that as k _0, (1+#* = e"

(c) Let X,xz" Xkbeii.d. Exp(1) , and let Mk max( X, Xc) (Maximu o Xh_X)_ Fiud the prohability deusity function of Mk:(Hint: P(min(Xh,Xz,Xs) > k) = P(Xi > k,x >k,X > k), how ahout max ?) (d) Show that k-0,the CDF o Z = Mk ln(k) is the CDF of Y. Hint: Note that as k _0, (1+#* = e"



Answers

Let $X$ be a nonnegative continuous random variable with cdf $F(x)$ and mean $E(X)$ .
(a) The definition of expected value is $E(X)=\int_{0}^{\infty} x f(x) d x .$ Replace the first $x$ inside the integral with $\int_{0}^{x} 1 d y$ to create a double integral expression for $E(X) .$ [The "order of integration" should be $d y d x . ]$
(b) Rearrange the order of integration, keeping track of the revised limits of integration, to show that
$$E(X)=\int_{0}^{\infty} \int_{y}^{\infty} f(x) d x d y$$
(c) Evaluate the $d x$ integral in (b) to show that $E(X)=\int_{0}^{\infty}[1-F(y)] d y .$ (This provides an alternate derivation of the formula established in Exercise $38 . )$
(d) Use the result of (c) to verify that the expected value of an exponentially distributed rv with parameter $\lambda$ is 1$/ \lambda$

All right. Now they have given it an expression for expectation effects. We're starting with the eye, but right now, so expectation off X in this particular problem is given as indicative. Go to infinity. This is X into F off X B X. So in the first month, their goal is to replace this ex tone with an expression. So I want to realize that part here. Someone to replace expert and giggle zero through X. This is one in tow, Levi. And if affects the extra meaning as it is. So if I heard this expression with proper arrangement of terms, the final expectation expression that I'm getting now is one into FX will give me FX. And now the order of indignation is due by the excellence is with the first part. It's not moving on to the second bark. They're told that we actually have to Now, in the order of this integration from, say, the vile e x to d x d vile. So I'm gonna force consider that given limits that we have here, you can see that the elements are made up functions off eggs, so including them. Equally wise, we have eyes from zero. Why is equal to X and outer limits? The homeowners X zero x equal Do infinity by turning point us on the next y axis. So I'm thinking this day that I'm here. It's an exact science. This is by access the sir prisons they questioned by Kordell zero two separate since the question X equal to zero. Okay. And we are viable to accept his lying passing Trojan. Okay, this is the origin. So no, we're gonna take a horse, understand? Because we're changing order to dig deeper if I take a strip like this. And if I shared that strip so we can see that from two starting from the line. Why quarterbacks? And here it is leading to infinity. And there's some lovely Why is going to be start from exactly Which is why I called you zero and this very cold. I hope you're infinity. So over here we can see that X is equal. Do. But on this side you can say that excess standing infinity. So the new limits that I will be getting now Okay, that will be first with I don't elements off X so excess from by tow X is equal to infinity on we have eyes going from zero. Why is equal to infinity? So if I needed with the new limits, the expectation of the expectation affects now can be two dinners so that limits of I will be writing art because their constant zero toe infinity and then I have integration excess from buy to infinity. The function remains as it is and we obtained order of indications of the X and B. Bye. So this is what Bill and Linda be part now moving on to the bar, See for the party. They have said that we're supposed to evaluate this inner integration which is actually the first or indignation with X. So if I even aware that I want to get a new expression for your fix so the outer limit remains as it is Now we know that small FX is the pdf and if I integrate this, we will get the CDF over so which is represented with a capital f affects. And so the indignation is done so I can put the limits from wired to infinity and do you so this part means, as it is now upon the limits I have. This is F off infinity minus f off riot. Anyway, Now we know that CD of the highest value possible is actually one because it ranges from 0 to 1. So we can therefore a place if often for knitting with one. So the final expression that I get as indictable zero to infinity in the bracket we have one minus f off. Why do you buy? So this is what is the expression for them Parsi now moving on to the next part for the Barney. There have been supposed to take the expression that we have derived Enciso, which is I just realized that you affects is integral zero to infinity black we have on minus the CDF off by and Dubai. Okay. Now they're told to consider the expansion the solution with parameter lambda and go take the CDF offered. So we can therefore write f off by the CD If the parameter lambda definition is one minus used to minus lambda Why? This is an advise ready than or equal to zero. I had zero in any other case since our indignation is from zero to infinity will be considering the first value so successful that in the expression off your eggs. Okay, so this is equal to individuality or to infinity. We have one minds and first apps. You'll devalue off for five. This is one minus is to minus lambda. Why? With diva? So we can cancel one here and minus and minus will become blessed. So I'm going to actually indignities to mine Islam there. Right. So you just minus lambda wise indeed. Nation upon minus lambda. And the limited strums you do infinitely. Not the nominee is a constant. I'm taking it outside in the upper limit for infinity. Here it is, two minus Infinity is zero minus E gs to zero gives me one. So therefore the final expression for expectation off X. When the parameters lambda it is given by one bite down. So we're done window. The part also

And this exercise for part A. We want to show that the statement is true and that's 40 is greater than zero. And the function F X is a pdf. So we're integrating over the interval from T to infinity. So for X in that interval, X is at least a biggest T, which means that X times ffx is at least a bigas tee times, eh, FedEx Therefore, if we integrate over this range, this is at least as big as this integral. So I'm just bringing the constant t outside of the integral here. Now this integral here is, by definition the probability that X is at least his biggest T. Which means that this integral probability that X is at least a biggest T is equal to one minus the cumulative distribution or the cumulative probability AT T. And so therefore, we have shown this Now for part B. We want to show the following and in the question were given the hint to Writeth e integral for the mean as the sum of two inter girls, one from zero to T and another from T to infinity. So if we follow that hand, the mean is defined this way. But in the question we are told that X is non negative. So the integral goes from zero to infinity. Now, using the hint, we write this as to into grills. And so from this we have into girl from T to infinity is equitably mean minus this integral. Now, using our result from part A, we can say that this is greater than or equal to three times one minus the cumulative distribution. So in party, we showed that this was greater than or equal to this. But these two things are equal. So we have this overall statement and from the question he is greater than zero. And the CDF is at most one which means that this is greater than or equal to zero now is t approaches infinity. This converges to zero So if we calculate the following limit and so the limit of this integral as T approaches infinity is the integral from zero to infinity and this is the expected value for X since our pdf is non negative and is only defined for X is at least zero. And so these to fax tell us that the limit as t approaches infinity of Tito my times one minus The cumulative probability at T must also equals zero

Everyone Radio going to solve problem number 25 here If if it's it was four in tow X minus one. The whole cube were one last memory Codex that's been recorded. So expectation off X is equal to integral mhm to the x f o fix the X. So here we had to using integration by parts there you, the vehicles you will be minus in the week. The you you will goto X it's the X. Then do you recall toe the X? We could talk if a fix f dissect because small left effects So expectation off X is equal to and they're a tow the x f l A fix, which is a good egg Scinto Captain left or thanks from a to B minus in the a b A full fix the X, which is ableto be f off the minus a f off a minus in the a Toby F or fix the X. So for a equals zero if off, Because what expectation off X is equal to the off one. Minus. Hey off zero minus a tow. The F 06 the It's Regis equals B minus A b. And for six, the X thank you

21. We consider the function of X to be the probability density function for a normal distribution for birdie. We want to show that this function is the maximum when X equals me. And to get the maximum value first, let's define B. Of X. For the normal distribution it equals one divided by σ Buoyed by Square Root of two. Boy blow it by E. There's a lot of minus X minus mu. Oil squared, divided by two, multiplied by Sigma Square. And to prove that P of X is maximum and X equals mu. We just Differentiate be relative to X. and equate the differentiation by the by zero. Then we get P dash of X equals the differentiation of E. Gives the and it's multiplied by the constant. We put the constant and we would E as it is. Then we want to blow it by the differentiation of the polynomial in the power. The economy in the power have a constant -1 divided by two segments square. Then we have power. We put the power then decreased one from the board and we must employ by the differentiation Of X, which is one. Then this is a p dash of x. We equated to zero and to get the value of X. We can remove the constance, then we have E. To the power of minus x minus mu. All squared divided by two sigma squared equals. Sorry, it's not equal. It's multiplied deployed boy, X minus mu equals zero. Here we have two options. The first option is to have it was about minus x minus mu. All squared away with two squared equals zero or x minus new equal zero. It is about uh minus any value. Can't be equal zero. This is a bust of function. It always gives about the value. Then we are sure that x minus mu equals zero, then X equals mu. This is the value of X. That makes B of X max. And to get the maximum body we just get P of new. We substitute by X equals meal in the function. Then we have one divided by sigma deployed by square root. Two boy are deployed by E. Was about of you because when we put you hear equals we put X equals here view. It gives it to the board of zero and it was about of zero is one. Then the maximum value is one divided by sigma deployed by square root of two boys. This is full body for Birdie. We want to show that B of X has points of inflection. Well X equals U plus sigma And x equals new -2. We can do this by getting the second derivative of P of X. B double digit of X And equated to zero. This gives us the points of inflection, this is part of p dash of x. Then we can differentiated again relative to x And equated to zero. Then the differentiation of E. The differentiation of two functions multiplied to each other. We take the first function E is about of minus x minus mu squared, divided by two sigma squared and we multiply it by the differentiation of the second one. The differentiation of x minus mu, give us one then plus we take the second function x minus mu and multiply it by the first function. The differentiation of the first function we differentiate E. The differentiation of E is just E. It's about minus x minus mu. All squared divided by two sigma squared. And we multiply it by the differentiation of the power which is minus one divided by two sigma squared, multiplied by two, multiplied by x minus mu. All this equal zero again we can get rid of the constant but here we have applause but we can get rid of E. Because we have said that in bertha it was about minus any value doesn't give us zero and it's always a bust of value then we can you move it and expressing it as a concept. Any value of X Can't get us zero value then what remains is one plus x minus mu deployed by minus one divided by two sigma squared, multiplied by two X minus meal. These all equal zero. We can take x minus mu and multiply it by x minus mu. Then we have one minus x minus mu squared multiplied. Boy we can cancel two with two. To apply it by one divided by sigma squared equals you. You can rearrange to get x minus mu All squared equals divided by sigma squared equals one. Then we can get x minus mu squared equals sigma squared. Then by taking the square root for both sides, you can get x minus mu equals Most of our -6, then we can get X equals you positive or minus sigma, And these are the two inflection points, one when X equals mu plus sigma, the other one when X equals u minus six.


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