ALL the other problems here work this same way. Consider the counts of offspring of given types as the sum of the 20 independent 1/0 ("indicator") random variables, one for each offspring.
2. (a) Two plants with wrinkled seeds, but known to be genotype Ww, are
crossed. Consider one offspring. Write X=1 if the offspring has wrinkled
seeds (P(X=1)=3/4) and X=0 otherwise. What are E(X) and V(X)?
(b) For 20 offspring, let Y be the number with wrinkled seeds.
Find E(Y) and V(Y).
3. (a) Two pink-flowered plants are crossed. Consider one
offspring. Write X=1 if the offspring has red flowers, and X=0 otherwise
(P(X=1)=1/4). Write Z=1 if the offspring has pink flowers, and Z=0
otherwise (P(Z=1)=1/2). Find cov(X,Z) and corr(X,Z).
(b) In a set of 20 offspring from this cross, let Y be the number with
red flowers, and T the number with pink flowers.
Find cov(Y,T) and corr(Y,T).
(c) There are Y with red flowers, and 20-Y with pink/white flowers.
What is cov(Y,20-Y)? What is corr(Y,20-Y) ?
4. OK, let's tidy up some notation -- I've run out of letters. In the cross described in the second half of the previous section, consider first one offspring and write
X1 =1, if it has pink flowers, wrinkled seeds, P(X1=1)=0.4 X2 =1, if it has pink flowers, round seeds, P(X2=1)=0.1 X3 =1, if it has white flowers, wrinkled seeds, P(X3=1)=0.1 X4 =1, if it has white flowers, round seeds, P(X4=1)=0.4In a set of 20 offspring, let
5. Returning to #3, consider again the 20 offspring plants,
each with P(red)=P(white)=1/4, P(pink)=1/2.
Let Y=number red, T=number pink, and W = number white (Y+T+W=20)
(a) What is the probability P(Y=y, T=t, W=w)? what is the name of this
probability mass function?
(b) What is the pmf P(Y=y) ? What is E(Y)? What is V(Y) ?
(c) Given that a plant is not white, what is the probability it is red?
Given there are 6 white plants, how many red ones do you expect?
(d) Given there are 6 white plants, what is the variance of the number of
red plants? Is it larger of smaller than the variance of Y found
in part (b)? Is this what you would expect?
(e) Given there are 6 white plants, what is the covariance of the number
of red ones and the number of pink ones. Compare this with your answers
to #3(b) and #3(c).
6. Here are the general formulae for means and variances in a binomial distribution, and for covariances in a multinomial distributions.
(i) Suppose X is B(n,p), so P(X=k) =
(n!/(k! (n-k)!) pk(1-p)n-k, k=0,1,2,3,....,n.
(a) For the case n=1, show E(X)=p, V(X)=p(1-p)
Then sum up the means and variances for the n independent trials, as in the
other examples, to get E(X)=np, V(X)=np(1-p)
(b) By summing k*P(X=k), show E(X)=np.
This method is no better than the other, and is a lot messier.
(c) By summing k*(k-1)*P(X=k), show E(X(X-1)) = n(n-1)p2,
and hence show V(X)=np(1-p).
(ii) Now suppose (X1,..., Xr) is multinomial Mn(n, (p1,...,pr)). This is the probability distribution we get when we have n independent objects, but they have r different possible types, and P(object is type j)=pj, j=1,...,r. We could use the pmf for this one also, but that gets really messy. Instead, we consider one object at a time, and sum.
(a) Consider just the first type. Then X1 is just B(n,p1).
Why is this? So now you can write down E(X1)
and V(X1). Also E(Xj) and V(Xj).
(b) Consider just one object (n=1). Show E(X1X2) =0,
and cov(X1, X2) = - p1p2.
What is cov(Xj, Xi) ?
(c) Now add up over n independent objects to show that
cov(Xj, Xi) = -n pipj.
(iii) Now suppose I know Xr=k.
How does this affect the prob dsn of
(X1,..., Xr-1) ?
(a)
Well, chucking away the k objects I know are type r,
now I have n-k independent objects, and
P(type j) = p*j = pj/(1-pr) (why?)
(b) So Xj is B(n-k, p*j) (why?).
So what is E(Xj | Xr = k) ? And what is
V(Xj | Xr = k) ?