Proof. By definition, equally likely events have equal
probability of happening. Suppose that the probabilty is p.
Since we are sure that something will happen, the total
probability of the events is equal to 1. Hence we have
Obviously p=1/n. Hence the probability of each event is
equal to...
I know that I need only to find a sing delta given an epsilon, but without knowing a, what good is knowing that |x - a| < 1?, for instance. What about considering the open interval (-∞, 0) U (0, ∞)? With this, I can show that for any x in I, the limit goes to zero, correct?
Could you elaborate...
|f(x) - L| = |f(x) - 0| = |f(x)| < ϵ.
Given the definition of the function, the only way this holds is if x is nonzero.
At this point I tried to consider 0 < |x - a| < δ. I'm trying to show that you can pick a δ such that i dunno...
If f(x) = 0 for all nonzero x and f(0) = some constant c, how can you show that lim = 0 for any x as x approaches a?
I tried using the definition of limit, but this is going nowhere.
So you use a matrix to solve the transition probabilities in the backwards equations? Can you elaborate on this. I am using Ross's book "Introduction to
Probability Models" and he doesn't explain how to do this, at least not in this section.
Consider two machines, both of which have an exponential lifetime with mean 1/λ. There is one repairman who can service machines at an exponential rate of μ.
How does one set up the Kolmogorov backward equations for such a scenario? I am not sure after finding the rates how to work those...
Consider two sequences, {a_n} and {b_n}.
If there is a one-to-one correspondence between these sets, can we conclude anything about their behavior considering, say, that we know that one is convergent?
Going further, can we conclude anything about the series resulting from these sequences?
Yes, although I still feel uncomfortable putting something down without knowing how to write a formula for it. I know that it isn't necessary, but it just seems a little more solid to me.
This sequence with subsequences converging to every natural number is definitely not as analogous as I thought it would be to the previous problem. I'll keep thinking about it, but if anyone has some helpful suggestions or hints then I would be glad to read them. Thank you for all the help thus far.