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4. If Xhas the following distribution;f() = B _" 'e Or derive E(Xk) assuming that k> -B.r>0...

Question

4. If Xhas the following distribution;f() = B _" 'e Or derive E(Xk) assuming that k> -B.r>0

4. If Xhas the following distribution; f() = B _" 'e Or derive E(Xk) assuming that k> -B. r>0



Answers

Show that $\mathrm{E}(X)=B-\int_{A}^{B} F(x) d x,$ where $F(x)$ is the cumulative distribution function for $X$ on $A \leq x \leq B$.

This question discusses proving the parameters for uniform distribution. Now we know that a uniform distribution has all of the values Given by one over B -A. Now we want to find A. You and in general when you want to find mu we can actually take the expectation of some value here, which we're going to call X. And that will allow us to find me right, it's expected values an average. Um So here we're actually going to take the integral from A to B. Of our distribution, which is you which is given by this value here. So this is going to be one over b minus A. And this is going to be a function of X, let's say D X. Now this is all a constant because our only variable here is X. So this just becomes X squared over to Times Our one Over B -A. Yeah, here in front. Now we can plug in for X here A. And B. And this gives us you plug in the B. And we plug in the A. And this will give us our values for a meal here. Now we can do the same thing but we're going to find prove the variance and the variance which I'm gonna general here by V is given by this following formula here, which is the expectation of X squared minus the expectation of X, that is quantity squared. Now we are resolved for the expectation here, right? This is the expectation of the value here. Now we can actually do the integral again plugging in from A to B. But this time we're actually going to go from one, this is just the value of the distribution here and this time we're gonna plug in an X squared here and integrate this out. Once you integrate this out, that will allow you to solve, you have now solved for this term here and then once you take this term and you square, that allows you to plug in for this value here. Once you plug in both of those values, that allows you to solve, and you should get the given answer for the variance of the uniform distribution Yeah.

Yeah. This problem with giving the moment generating function is equal to 1/6. Yeah. Either the T plus 26 Either the two T. Lost 36 Each of the three team, we would like to find first the expected by now the expected value is equal to the derivative of the moment generating function evaluated at zero. And so if we want to find the expected value we'll find the moment generating function And then plug in zero. Okay. And so the derivative of the moment generating function. There's one set, you know the tea plus four six. You know the two T past 96 You know the three teams and then plugging in zero here Gives us 1/6 eat of the zero. Lost 460 last 96 You to zero Which was 14 6 which is 7/3. And so the expected value of why Is 7/3? Yeah, for beef. We want to find the variance now. Before and find the variance. We need to find the expected value of Y script. The expected valuable. I swear it is the second derivative of the moment generating function evaluated at zero. Our second derivative there's one sex either the T Past 8, 6 either the two T Plus 27 6. Each of the three T. No parking. And zero. We have em double prime zero. It was 1 6- zero. Was 8/6 You to the zero was 27 6, even the zero. Mhm. And so I did all this up. We get 36 6 which is six. Remember that's just the expected value of Y squared But we want the variants, the variance of why is the expected value of Y squared minus the expected value of why? I swear. And so this is six squared minus 7/3 square. And so this is 36 49 over nine and 36 -49/9 will give us five nights for our variants now, for our last one, Yeah. We have been given at the moment generating function. Yeah. Is 1/6 of the T. Was to search either the two T Plus 3, 6 each of the three teams. And we want to find the probability distribution function. Mhm. Now let's patch this up your match up the X opponent with the probabilities. We have a one T In 16. This means the probability of why is 16 If Y is one, Yeah. And the next part we have to six and 2. So that means it's 26 or 1/3 if y is too. And here lastly we have 36 and three until this means we have 36. If Y is three, this is our distribution function.


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