A mother and a father have $6$ children. The $8$ heights in the family (in inches) are $N (µ, σ_2)$ r.v.s (with the same distribution, but not necessarily independent).
(a) Assume for this part that the heights are all independent. On average, how many of the children are taller than both parents?
(b) Let $X_1$ be the height of the mother, $X_2$ be the height of the father, and $Y_1,...,Y_6$ be the heights of the children. Suppose that $(X_1, X_2, Y_1,...,Y_6)$ is Multivariate Normal, with $N (µ, σ_2)$ marginals and $Corr(X_1, Y_j ) =r$ for $1 \le j \le 6$, with $r < 1$. On average,how many of the children are more than $1$ inch taller than their mother?
I did so far the following:
a) $P(I=1)$ if height is higher and $0$ otherwise. $P(I=1)=P(Y_i \ge x_1)P(Y_i \ge x_2)=[1-Φ(x_1)][1-Φ(x_2)]=E[I]$
b) $P(I=1)=P(Y_i \ge x_1+1)$
I don't know if my reasoning is correct and even if it is what to do next?
For part a), the probability that any given child is taller than both parents is $\frac13$.
Reasoning: The three heights are from identical normal distributions. Thus there is an equal chance that any of the three heights is the tallest height.
So the expected number of children taller than both parents is $6\cdot\frac13=2$ (by linearity of expectation).
For part b), take any one child. Let its height in inches be $C$. Let the mother's height in inches be $M$. Let $D$ be the difference $C-M$. We are interested in $C-M>1$.
But $C-M$ is normal with mean $0$ and variance $2\sigma^2$.
From this you should be able to figure out the probability that $C-M>1$.
Once you have the probability for one child, you can use linearity of expectation to find the expected number of children who are at least one inch taller than Mom.