Choose a number X at random from the set of numbers \{1,\ 2,\ 3,\ 4

quiquenobi2v6 2021-12-14 Answered
Choose a number X at random from the set of numbers {1, 2, 3, 4, 5}. Now choose a number at random from the subset no larger than X, that is, from {1, , X}. Call this second number Y.
a) Find the joint mass function of X and Y.
b) Find the conditional mass function of X given that Y=i. Do it for i=1, 2, 3, 4, 5.
с) Are X and Y independent? Why?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

twineg4
Answered 2021-12-15 Author has 33 answers
We are given that XDunif(1, ,5). Observe that YX almost certainly and that also Yϵ{1, , 5}
(a)
For kl we have that
P(Y=k,X=l)=P(Y=hX=I)P(X=I)=1l15.
I tk>l, we have that P(Y=k,X=l)=0
(b)
Using the law of the total probability, we have that
P(Y=i)=l=i5P(y=i,x=i)=l=i515l
Now,for i we have that
P(X=iY=i)=P(Y=i,X=i)P(Y=i)=15l{i=1}515l
(c)
No, these random variables are not independent Since YX, take any k>l, for example take k=2, l=1.
Hence P(Y=k,X=1)=0. On the other hand, it is pretty obvious that P(Y=k)>O and P(X=1)>0.
Thus
P(Y=k,X=1)P(Y=k)P(X=I)
so they can not be independent.

We have step-by-step solutions for your answer!

Thomas White
Answered 2021-12-16 Author has 40 answers
For X, we have 1 chance out of five to get any of the five numbers, so
Px(k)=15=0.4 for any integer k between 1 and 5.
For Y,we can use the theorem of total probability, because we can divide Y
in five different cases depending on how it went with the value of X. We can
remove impossible cases (of probability 0), those cases happen when X is
smaller than the value we want to get with Y.
Py(1)=P(Y=1)=P(Y=1X=1)×P(X=1)+
P(Y=1X=2)×P(X=2)+P(Y=1X=3)×P(X=3)+
P(Y=1X=4)×P(X=4)+P(Y=1X=5)P(X=5)=
0.1×0.2+0.5×0.2+13×0.2+0.25×0.2+0.2×0.2=0.45666
Py(2)=P(Y=2)=P(Y=2X=2)×P(X=2)+
P(Y=2X=3)×P(X=3)+P(Y=2X=4)×P(X=4)+
P(Y=2X=5)×P(X=5)=0.5×0.2+13×0.20.25×0.2×0.2×0.2=0.25666
Py(3)=P(Y=3)=P(Y=3X=3)×P(X=3)+
P(Y=3X=4)×P(X=4)+P(Y=3X=5)×P(X=5)
=13×0.2+0.25×0.2+0.2×0.2=0.15666
Py(4)=P(Y=4)=P(Y=4X=4)×P(X=4)+
P(Y=4X=5)×P(X=5)=0.25×0.2+0.2×0.2=0.09
Py(5)=P(Y=

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions