1890 Shares

Statistics: Conditional distributions. Any help?

Statistics: Conditional distributions. Any help? Topic: Help me with my statistics homework
July 16, 2019 / By Delight
Question: This isn't a homework question; I have a final in my statistics class tomorrow, and I need to understand conditional probability. :( Let X and Y be independent and have the same geometric distribution with success probability p. Find the conditional distribution of X given X + Y = n. The book doesn't seem to give a good explanation of this, so any help would be great! Thanks!
Best Answer

Best Answers: Statistics: Conditional distributions. Any help?

Bryanne Bryanne | 5 days ago
Hope this isn't too late. I'm going to assume that the geometric distribution you have is for the number of failures before you get 1 success. That is, Pr(X=j) = p(1-p)^j One of the features of the geometric distribution is that it has no memory meaning that the conditional probability does not depend on the number of prior failed trials. Using Bayes Theorem Pr(X=j|X+Y=n) = Pr(X+Y=n|X=j)Pr(X=j)/Pr(X+Y=n) Pr(X=j|X+Y=n) = Pr(Y=n-j)Pr(X=j)/Pr(X+Y=n) Pr(X=j|X+Y=n) = p(1-p)^(n-j) p(1-p)^j / Pr(X+Y=n) Now Z = X+Y is, I believe, a negative binomial random variable with parameters 2 and p. Pr(Z=n;r=2) = Γ(n+2)/[n!Γ(2)] p²(1-p)^n Since Γ(m) = (m-1)! then Γ(n+2)/n! = (n+1)!/n! = n+1 So Pr(Z=n;r=2) reduces to p²(n+1)(1-p)^n Substituting Pr(X=j|X+Y=n) = p(1-p)^(n-j) p(1-p)^j / Pr(X+Y=n) Pr(X=j|X+Y=n) = p(1-p)^(n-j) p(1-p)^j / [p²(n+1)(1-p)^n] Pr(X=j|X+Y=n) = 1/(n+1) I can't guarantee that this is correct, but I believe I was correct when I stated Z=X+Y is negative binomial, and that my use of Bayes theorem was correct. Good luck.
👍 258 | 👎 5
Did you like the answer? Statistics: Conditional distributions. Any help? Share with your friends

We found more questions related to the topic: Help me with my statistics homework


Bryanne Originally Answered: Normal Distributions Statistics?
Dear Ambie2992, In R, the name of the normal cumulative distribution function is "pnorm" (the cumulative distribution is what you want to calculate the probabilities in your question). By default, pnorm uses the standard normal distribution (with mean = 0 and sd = 1), but this function allows you to optionally give other values for the mean and standard deviation if you ever want to work with the normal distribution in other forms. For your questions, pnorm(x) will give you the probability that Z is less than or equal to x, that is, P[Z ≤ x]. (a) P[Z < 1] = P[Z ≤ 1] (since the normal distribution is continuous), so the calculation in R of P[Z < 1] is simply pnorm(1) which evaluates to 0.8413447 (to seven decimal places). (b) You can calculate P[Z > 0.25] two ways. The first way is to recognize that P[Z > 0.25] = 1 - P[Z ≤ 0.25], so in R you would write 1 - pnorm(0.25) which evaluates to 0.4012937 (to seven decimal places). However, there is another optional parameter that you can use with the pnorm function that allows you to change the evaluation from the default of the lower tail (or left tail) to the upper tail (or right tail). In this situation, pnorm(x, lower.tail = FALSE) will give you the probability that Z is greater than x, that is, P[Z > x]. Thus, by using lower.tail = FALSE in your evaluation of pnorm, you get the same result, but don't need to subtract the function evaluation from 1. pnorm(0.25, lower.tail = FALSE) evaluates to 0.4012937 (to seven decimal places). (c) For P[-1 < Z < 1] (= P[-1 < Z ≤ 1], since the normal distribution is continuous), it helps to look at a graph if you need more clarity on its meaning. You should easily recognize that P[-1 < Z ≤ 1] = P[Z ≤ 1] - P[Z ≤ -1], so in R you merely use pnorm(1) - pnorm(-1) which evaluates to 0.6826895 (to seven decimal places). I hope that my explanations have given you what you need to know, but if you want further help, please let me know. Best wishes in your studies!

Alleen Alleen
Suppose X+Y=Z. Then Z has negative binomial with parameters (r=2, p=p). According to the Bayes theorem, you need to find P(X=n-y)/P(Z=n). P(X=n-y)=p*(1-p)^(n-y) P(Z=n)=Г(2+n)*p^2*(1-p)^n / (n!*Г(2)). Г(2)=1!=1 notice that Г(2+n)/n!=(1+n)!/n!=n+1 After the division, the equation becomes 1/ [(n+1)*p*(1-p)^y]. That's your distribution.
👍 110 | 👎 -1

Alleen Originally Answered: Statistics, confidence intervals, distributions?
The formula for a 95% confidence interval with a large sample size is: x_bar (+/-) [z_(alpha/2) * (standard deviation/sqrt(n)) So, A. 140 (+/-) 1.96 * (25/sqrt(200)) 140 (+/-) 1.96 * (1.77) = 140 (+/-) 3.47 [136.53, 143.47] B. 130 (+/-) 3.47 = [126.53, 133.47] No, 140 is not contained in this interval, so you can not conclude an association exists between glaucoma and blood pressure. Hope that helps!
Alleen Originally Answered: Statistics, confidence intervals, distributions?
It skill that for the time of accordance on your 6-merchandise pattern, the genuine underlying recommend of the distribution that generated the information has a ninety 5% danger of falling between 86.14 and ninety seven.86. i'm uncertain what meaning in terms of reading your effect, in view which you haven't any longer defined the hypothesis you're checking out. Statistical information are not recommended with out a context wherein to word them.

If you have your own answer to the question help me with my statistics homework, then you can write your own version, using the form below for an extended answer.