Asked by JuanPro
A fair coin is flipped independently until the first Heads is observed. Let K be the number of Tails observed before the first Heads (note that K is a random variable). For k=0,1,2,…,K, let Xk be a continuous random variable that is uniform over the interval [0,3]. The Xk's are independent of one another and of the coin flips. Let the random variable X be defined as the sum of all the Xk's generated before the first Heads. That is, X=∑Kk=0Xk. Find the mean and variance of X. You may use the fact that the mean and variance of a geometric random variable with parameter p are 1/p and (1−p)/p2, respectively.
E[X]= - unanswered
var(X)=
E[X]= - unanswered
var(X)=
Answers
Answered by
kadu
I ma stuck with the same problem, somebody please give us a hint.
Answered by
RVE
E[X]= 3
var(X)= 6
if you are reading this answer, please help us with the whole problem set....
var(X)= 6
if you are reading this answer, please help us with the whole problem set....
Answered by
Gilraed
The E[Y] is equal to E[N]*E[X], being Y=X1+...+Xn. Now, we know that N is the number of coin flips, and Xk is independent and uniformly distributed between 0 and 3. This means that E[X]=3*1/2=3/2, with a variance of 3^2/12=3/4.
Now, for N we are given 1/p and the variance. Since it's a fair coin, p=1/2, and so E[n]=2 and var[N]=2.
With this, E[Y]=2*3/2=3
var(Y)=E[N]*var(X)+(E[X])^2*var(N)=2*3/2+2^2*3/4=3+3=6
Hope it's understandable.
Now, for N we are given 1/p and the variance. Since it's a fair coin, p=1/2, and so E[n]=2 and var[N]=2.
With this, E[Y]=2*3/2=3
var(Y)=E[N]*var(X)+(E[X])^2*var(N)=2*3/2+2^2*3/4=3+3=6
Hope it's understandable.
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