How to calculate the correlation between the number of heads of 100 toss of coin and the number of heads of the first 10 toss of those 100 tosses? It is a bit cumbersome to explain:
Toss a coin is a Bernoulli distribution, with the probability of seeing a head is p
if we toss this coin 100 times, we should expect $X_{1}$ times of head. Within that 100 toss (this is important, we are NOT tossing another 10 times), we should see $X_{2}$ heads from the first 10 toss.
How to calculate $corr(X_{1}, X_{2})$ ?
The only thing I can think of is, $X_{1} > X_{2}$, practially, we are doing two sets:

*

*toss a coin 10 times, we see $X_{2}$ heads

*independently toss a coin 90 times, we see $X_{3}$ heads

we want to calculate $corr(X_{2}, X_{2} + X_{3})$
 A: Let $Y_i$ take value $1$ if toss $i$ gives heads and let it take value $0$ otherwise.
Then to be found is:$$\mathsf{Corr}\left(\sum_{i=1}^{10}Y_i,\sum_{j=1}^{100}Y_j\right)=\frac{\mathsf{Cov}\left(\sum_{i=1}^{10}Y_i,\sum_{j=1}^{100}Y_j\right)}{\sqrt{\mathsf{Var}(\sum_{i=1}^{10}Y_i)}\sqrt{\mathsf{Var}(\sum_{i=1}^{100}Y_i)}}$$
Note that on base of bilinearity of covariance, independence and symmetry: $$\mathsf{Cov}\left(\sum_{i=1}^{10}Y_i,\sum_{j=1}^{100}Y_j\right)=\sum_{i=1}^{10}\sum_{j=1}^{100}\mathsf{Cov}(Y_i,Y_j)=10\mathsf{Cov}(Y_1,Y_1)=10\mathsf{Var}Y_1$$
Further, also on base of independence and symmetry: $$\mathsf{Var}\left(\sum_{i=1}^{10}Y_i\right)=10\mathsf{Var}Y_1$$and: $$\mathsf{Var}\left(\sum_{i=1}^{100}Y_i\right)=100\mathsf{Var}Y_1$$
Leading to answer: $$\frac1{\sqrt{10}}$$
It is not even necessary to calculate $\mathsf{Var}Y_1$.
A: Your splitting $X_1$ into $X_2$ and an independent $Y_3$ is a correct approach and you get $$\textrm{Cov}(X_{2}, X_{1})=\textrm{Cov}(X_{2}, X_{2} + X_{3})=\textrm{Var}(X_2)+0=10p(1-p)$$
So $$\textrm{Corr}(X_{2}, X_{1})=\frac{\textrm{Cov}(X_{2}, X_{1})}{\sqrt{\textrm{Var}(X_2)\textrm{Var}(X_1)}}=\frac{10p(1-p)}{\sqrt{10p(1-p)100p(1-p)}} = \frac{1}{\sqrt{10}}$$ as drhab found
