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12. Let the d.f. have density that - continuous and positive on am interval such that F(b) _ F(a) < 0. (The results are fact valid if we only suppose that F is c...

Question

12. Let the d.f. have density that - continuous and positive on am interval such that F(b) _ F(a) < 0. (The results are fact valid if we only suppose that F is continuous. )(a) Show that if X has density then Y = F(T) is uniformly distributed on (0. 1). (b) Show that if U u(O. 1) . then F-!(U) has density

12. Let the d.f. have density that - continuous and positive on am interval such that F(b) _ F(a) < 0. (The results are fact valid if we only suppose that F is continuous. ) (a) Show that if X has density then Y = F(T) is uniformly distributed on (0. 1). (b) Show that if U u(O. 1) . then F-!(U) has density



Answers

Show that the function is a probability density function on the show that the function is a probability density function on the specified interval. $$f(x)=\frac{12-x}{72} ;(0 \leq x \leq 12)$$

For this problem, we want to show that F of X which is equal to this piece wise function is a probability density function on the interval negative infinity to infinity. RK is greater than zero in here. We define have a bex to be a probability identity function. If the integral from ATB of F of XDX is equal to one. No, the integral from negative infinity to infinity of F of X dx. This is equal to for this case, since this is speech wise we right the integral into the integral From negative infinity to zero of F of X. Dx Plus the integral from 0 to infinity of F of X Dx. And based on description if access less than zero, this is equal to zero. So we have zero here plus The integral from 0 to infinity of K times erase negative K X dx. So since we have and infinite down then we have to change that to a variable let's say T. And you will have limits as the approaches infinity Of the integral from 0 to T of K times erase negative K X D X. And then you will do substitution here. We want to let U equal to negative K X. This gives us do you equal to negative KD X which gives us negative one over Kdu equal to dx. So if x is equal to zero, this gives us you equal to zero. And if X is equal to T this gives you equal to negative Katie. So then this would be limit as the approaches infinity. Ah the integral from zero to negative Katie of K times erase to you, Times -1 over Kdu. This is equal to negative limit as T approaches infinity of the integral from zero to negative Katie of the race to you. D you integrating This gives us negative times to limit as T approaches infinity Of the race to you evaluated from 0 to Negative Katie. And if you is negative Katie, this is equal to negative limit. Sc approaches infinity of the race to negative Katie Minour the race to zero And erases your as one. So evaluating this limit, we get negative of the race to negative infinity -1 And this is equal to negative of one over erase infinity -1. We know one over erase infinity approaches zero. So this is equal to one. And we have shown that f of X is indeed a probability density function.

So the problem didn't see function is going to be when my eggs or they endured one comma a So the property to stairs stairs area under the function If affects or the interval should be equal to one that is the individual ffx DX want to eat should be equal to one. So let's capitals integral. Yeah, yeah. So this satisfies the second property.

This function given here would not be a probability density function because, um, from excuse me from X is 1 to 2. They would take on negative values and f of X must always be positive. And since here it would be less than zero. This disqualifies it since we can't have negative probabilities and therefore this would not be a probability density function.

Problem line. We want to show that this function is a probability density function for the terrible of X between zero and infant. We have to check two things. The first thing is to check F of X is always a positive value. Exhale is always about the body. We can see that X is between zero and infinity. Then X is always positive. And we have F of X equals all those opposed to value and a square root which gives always about the value. Then every X is always positive and it's a bust of function. A busted function. The second chick is to make sure the area which is integration for the dynasty function through the in terrible from zero to infinity equals warm. We don't know if it equals one or not. This is a second check. Let's get the integral. This is an improver integrate. Then we take the limit for the integral from zero to be for F of X. Which is the point as X divided by X squared plus one to the power of one half. And the limit when beat in this to infinite the X. We can see that if we must blow it the nominated by two and we divided by two. We have the differentiation of the denominator. In the denominator the differentiation of the denominator is two. X. We have it here and denominator then we can add one to the power. Then divide by the new beau. The power here is minus three divided by two. Then we add want to the power then divide by the new book equals half. Multiplied by the limit of the integration gives X squared plus one. We add one to the power we have minus half. Then divide by minus off. We substitute from zero to be. We substitute by the upper limit. First the upper limit when x equals speed. We can remove one half with one half and you take only minus. We have p squared plus one. It was about uh minus half minus. We substitute by X equals zero gets warm. Was about negative health. By taking the limit for this term. We can see that we have one divided by the square root of infinity squared plus one, which is one divided by infinity, which gives zero. Then it's minus zero minus one to the power of minus half gives one. Then the final answer is one. And this is correct because we want the integral equals one. Then the second check is that fight and this is a dynasty function, probability density function.


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