Hi all, Just two things on MGFS please 1) is there a requirement that all MGFs converge as x->infty? this is implied by ch3 q7, although i have not seen this in the course notes. in that case does the denominator of all fractions need to be positive? what would this mean if there was no denominator, how would we find the range for t? 2) question 9 is really puzzling me. i cant get to the answer at all! i attempted matching the expansion, i got (0.5*sigma^2*t^2)^4 / 4! to match the term t^4/4! E[(X-mu)^4] so this implies the answer is (1/16 sigma^8 t^8)/t^4. this isnt even close to the answer of 3simga^4. can you see what i am doing wrong? i also tried the alternative method of expanding E[(X-mu)^4] to obtain e(X^4)-4e(X^3)mu+6,uE(X^2)-3mu^4 and putting in each moment, but the MGF of the normal dist is (e^mut+0.5sigma^2t^2) the differential of this is mu+1/2sigma^2*t so when t=0, E(X) becomes just mu, E(X^2) is mu^2 etc so the expansion is literally all just mu's and they all cancel out. Again, i must be doing this wrong? Really sorry this is a long request, please could you let me know where i am going wrong with each method please?
Just to add to this, am looking at Ch 3 Question 2, and the range given is the denominator -1<denominator<1, so doesnt fit in with what i had previously thought that the deonimator must be positive for MGFS, is there a rule that im missing? Thanks
I don't have the course notes, but there is no requirement that MGFs converge. There isn't even a requirement for an MGF to exist
Ah great thank you !! Do you have any ideas why a condition that the fraction must be positive would be imposed?
I'm afraid not, as I'm not sure which fraction you're talking about or the context or the question. Happy to help if you type out more details.
oh that would be great thank you! so part a is to find the MGF of f(x)=ke^-(2x), where x>R i have found this correctly as MGF= [ke^(R(t-2))]/[-t+2] the bit i am confused on is part b, " state the values of t for which the formula in part Ia is valued" the solution is "the integral converges as x->infinity only when -t+2 is positive, therefore we require t<2. Im just confused why we require the integral to converge as x-> infinity and whether this question will only come up when the MGF is a fraction. Thanks so much in advance
We say that an MGF for X exists if E[exp(tX)] converges for t in some neighbourhood of zero. i.e. there exists some (potentially very small) number y such E[exp(tX)] converges for all values of t between -y and y. When part a says 'state the values for which the formula is valued' I think it means 'state the values of t for which the MGF exists. So it's asking for which values of t the integral from R to infinity of the below converges: exp(tx)*k*exp(-2x) = exp((t-2)x)*k We can ignore the k because it's just a constant multiplier, and focus on when exp((t-2)x) will converge. Let's simplify by substituting u=t-2, and ask ourselves when does the integral from some constant R to infinity of exp(ux) converge? If you draw the graph of exp(ux) for u>0 you can see it goes up to infinity as x goes to infinity. So clearly the integral of it from any constant R up to infinity diverges. When u=0, exp(ux)=1 and again hopefully you can understand the integral from any constant R to infinity of 1 will not converge. When u<0, exp(ux) is bounded above by exp(uR), because exp(ux) is monotone decreasing. And it decays fast enough that the integral from R to infinity converges rather than diverges. So what we need for our integral to converge is u < 0. In other words, we need t < 2. An integral does not need to have a solution which is a fraction in order for it not to converge. Hopefully the way I have discussed the above explains this.
When I said “We can ignore the k because it's just a constant multiplier, and focus on when exp((t-2)x) will converge.” above I meant the integral of exp((t-2)x), to be clear.
Cant thank you enough for the detail you went into here, this is so useful and its really cleared this all up for me! Thank you so much!!!!
