Let Y_1, Y_2,...Y_n denote a random sample from the density function given by



Find a sufficient statistic for theta. pretty sure I got this(sum of the y's from i = 1 to i = n)

Find the Maximum Likelihood estimator of theta. Tried taking the derivative of the likelihood function and setting it equal to 0. I used product rule but got the same term for both parts and could not figure out how to set it equal to 0. I'm not sure whether I screwed up the derivative or whether I just can't see it.

Is there a better way to find the MLE for this problem?

You need to write the density for fn(y|theta) by multiplying n densities for the y1,y2,....yn. Then to find a sufficient statistic, use the factorization critereon. That is, identify one factor of fn(y|theta) that can depend on only the y's but not theta, and the other factor that depends on theta but only depends on the y's through the sufficient statistic r(y) where y = y1,y2,..yn:

fn(y|theta) = u(y)*v(theta,r(y)).

Can you do that if the sufficient statistic r(y) is the sum of the y's, or does it need to be something else?


You want to maximize the fn(y|theta) that you got above from multiplying the n densities. But maximizing fn(y|theta) is the same as maximizing the log[fn(y|theta)], and that makes it easier. Remember that a log of a product is a sum of logs, and log(1/theta^n) = -n*log(theta). The part that doesn't depend on theta just becomes an added log term that goes to zero when you take the derivative.

Just got introduced to sequences and am having bit of trouble figuring out how to systematically decide how to write the nth term.

given the sequence {1, -2/3, 4/9, -8/27...} I can clearly see a pattern that can be describe as follows: Starting from 1/1, the denominator of each successive number in the sequence is 3x the denominator from the previous number in the sequence. The numerator appears to be 2x the previous number in the sequence's numerator, alternating between positive and negative.

I don't know how to say that generally though. When I try to describe each number in terms of its respective n, looking for a pattern, the obvious pattern disappears, which leads me to believe I'm thinking about it wrong. For example:

a_1=n
a_2=-n/(n+1)
a_3=(n+1)/3n
a_4=-2n/(7n-1)

as you can see, no real pattern is apparent there, and in addition, it seems to me, that each number could be described in terms of its specific n in a lot of different ways, so this method doesn't seem productive. What would be a better way to look at these patterns?

What do you multiply the last by to get the next?

-2/3 * whatever number corresponds with the previous n.

so would you say something like

a_n=-2f_(n-1)/3 ?

I'd say:

a_n = (-2/3) * a_{n-1}

But that's not so pretty, since it requires knowing a_{n-1}

Notice that we can write the terms as:

a_0 = 1
a_1 = 1 * (-2/3)
a_2 = 1 * (-2/3) * (-2/3)
a_3 = 1 * (-2/3) * (-2/3) * (-2/3)
...

Okay ya, so a_n = (-2/3)^n.

As a convention do we always say that the first term of a sequence is the a_0 term?

edit: maybe it's better to say a_n=(-2/3)^(n-1)? This way we can start as a_1=1 and it's not confusing (to noobs like me).

You can define sequences that start at 0, start at 1, start at 108, whatever.

When you're given a sequence, though, you don't get to pick.

If they tell you a_0 = 1, whatever you write down had damn well better equal 1 when you plug in 0.

If they give you a sequence and say to find a "general term," as in this question, you should write a_n = <blah> for n >= <starting_val>

So a_n = (-2/3)^n for n>= 0.

Well, got an 84 on that signals exam...kind of meh but the average was a 65 with a standard deviation of 20, so not too bad I guess. Thanks to Bruce and others for helping me, would have gotten lower I'm sure without you guys helping.

Shouldn't this be in the BBV forum? Lol nah jokes. Well done buddy.

Hey, belated thanks for this. Quite an obvious oversight in the end!

for the sequence a_n= 3^(n+2)/5^n I am having a hard time seeing, analytically, why the limit is 0.

(ignoring the fact that L'Hospital is meant for functions)

I see that the general equation for this sequence has an indeterminate form of type inf/inf and so I apply L'Hospital but that only yields another indeterminate form of inf/inf, which, unless I am making a glaring oversight, seems to only produce another inf/inf for each subsequent application of L'Hospital's rule (or at least, applying it a few times gives a noob the impression that that is the case). This would cause me to think it goes to infinity.

If the lim is 0 that means that either the numerator goes to 0 while the denominator goes to some finite value, which clearly isn't the case, or the denominator, at some point, exponentially gets greater than numerator.

What method should I be using to evaluate the limit, or what am I doing wrong?

I suppose if I think of a_n as 9*(3^n/5^n) it becomes more apparent that the ratio between the exponential part of the equation is such that the denominator is going to get exponentially greater, but even still, for some reason my intuition tells me that

3^n/5^n is the same ratio as (3*3*3*3*3*3*3*3...../5*5*5*5*5*5*5*5...)
Of course that's not correct but might explain why it confused me.

Still, whats the best way to evaluate the limit?

edit: i see we posted at the same time.

It's just 9*(3/5)^n. When n gets very large, any positive number less than 1 will approach 0 when you multiply it by itself n times.

I'm not great at these, but for all n larger than 4, the numerator is going to be smaller, and increasingly smaller than the denominator. That's not an elegant proof, but intuitively it seems clear that the limit is zero.

!!! Dangit. That's right. I just learned that too.

R^n approaches 0 if R < 1 and approaches infinity if R > 1

for the sequence a_n= ((-1)^n*n^3)/(n^3+2n^2+1)

I can show that the sequence is bound below by -1 and above by 1, if I could show that it is not monotonic I could then say that it certainly is divergent.

We haven't learned about mathematical induction yet (though I've looked into it a bit), but would showing that it is not monotonic via mathematical induction require using the same type of analysis I would perform using differential calculus? In other words, if I was to show that the first derivative of a_n undergoes multiple sign changes I would have also shown that a_n is not monotonic, would proving this inductively use the same method?

It's obvious that it's not monotonic since each positive term is followed by a negative term, and each negative term is followed by a positive term. Induction requires 2 things. First show that there is some value N for which a_N is positive. Second, show that if a_n is positive, then a_(n+2) is positive. Together that proves that a_(N+2k) is positive for all integer k > 0. Then show that a_(N+1) is negative, and that if a_(n+1) is negative, then a_(n+3) is negative. Together that means that a_(N+2k+1) is negative for all integer k > 0. So A(N+2k) > A(N+2k+1) < A(N+2k+2) for all k. Since monotonicity requires that A(N+2k) >= A(N+2k+1) >= A(N+2k+2) or A(N+2k) <= A(N+2k+1) <= A(N+2k+2) for all k, a_n is not monotonic.

BTW, just being bounded and non-monotonic doesn't mean it's divergent. For example, (-1)^n * e^-n is bounded above by 1 and below by -1, and it is non-monotonic, but it converges to 0. The above sequence was divergent because there was no single number which the sequence would become arbitrarily close to for all n sufficiently large.

Yes. That makes sense. Denying the antecedent on my part.
If it's bounded & monotonic then it converges doesn't imply that if it's not monotonic then it wont converge.

Unrelated:

I could figure this out the long way, but I'm trying to think of what tools I have that can answer a simple question like the following:

20 minutes separated into alternating intervals of 1 minute (R) and 1.5 minutes (J). What sort of math is useful for solving this type of problem?

What's the question? The number of intervals? There are 20/2.5 = 8 pairs, or 16 intervals.

I think its probably helpful to know what you're asking. Essentially it would be alternating between running and walking for 20 minutes, wanting to know how long you'd do each.

I just realized that you could divide 20 by 2.5, which would constitute 8 "blocks" of running/walking, which means you'll do each 8 time and then you just multiply by the respective amount to find out how much of each you'll be doing.

I'm starting to see that half the battle is knowing what kind of question you are asking. I found myself thinking "can i use some sort of calculus optimization on this?" Also, when thinking about the problem where we are clearly not monotonic since the (-1) would alternate between positive and negative; So I knew A way to figure out whether it was monotonic or not (1st derivative test etc, even though, as you pointed out, that in itself doesn't mean it doesn't converge), but there was a much more obvious way, as the alternating 1, -1, would show. Is it laziness that doesn't allow me to see that, or is it a lack of exposure to enough problems so I'm not asking the right questions of myself when I'm trying to problem solve?

Want to know the limit as n gets bigger for the sequence a_n=(-3)^n/n!

Is the following acceptable:

If lim as n gets bigger of abs(a_n)=0 then so does a_n

abs((-3)^n/n!) = 3^x/n!

plugging in a few values

a_1=3
a_2= (3*3)/(2*1)
a_3=(3*3*3)/(3*2*1)
a_4=(3*3*3*3)/(4*3*2*1)
if we rearrange a_4 a bit
a_4=(3/4)[(3*3*3)/(3*2*1)]
which, in a hand wavy general kind of way implies
a_n=(3/n)*(some thing that is just a number)

Now if we take the limit as n gets bigger (3/n) is going to go to zero and, since the other thing is just a number and 0*any number=0, we've shown that abs(a_n) goes to zero as n gets bigger and as a result we know that a_n does too.

You're close. This part:

which, in a hand wavy general kind of way implies
a_n=(3/n)*(some thing that is just a number)

isn't really good enough. BUT, note that the "thing that is just a number" is actually a_(n-1). Now you have a_n = -(3/n) * a_(n-1) (you've also lost the minus sign here for some reason?)

Now it should be super simple to do the ratio test.

I lost the minus sign because I'm actually taking the limit of the absolute value of the sequence. I'm under the impression that (I think because of the squeeze theorem) if I can show that the abs value of the sequence has a limiting value of zero then we know that the original sequence does as well.

I am a bit confused though. I buy that the "just a number" part is a_(n-1) but I just see that as 3^(n-1)/(n-1)!, which looks, to me, no simpler than the original form of a_n. I thought if I could pull out that leading (3/n) term, a term I understand how to take the limit of, I would be able to say something about the whole sequence, but it seems not. Rather, I am just back to taking the limit of a sequence I don't really understand how to take the limit of (maybe it's the factorial that is throwing me off?). What am I missing?