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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:

  1. toss a coin 10 times, we see $X_{2}$ heads
  2. independently toss a coin 90 times, we see $X_{3}$ heads

we want to calculate $corr(X_{2}, X_{2} + X_{3})$

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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 $\mathsf{Var}Y_1$.

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  • $\begingroup$ Goedenmiddag , thanks a lot for answering this. however, I believe, it is not correct to model Yi and Yj as independent. We are trying to find the correlation between "the first 10 tosses" and " all the 100 tosses from which the first 10 tosses are chosen". These 2 random variables are natually dependent $\endgroup$
    – user152503
    Dec 24 '20 at 15:24
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    $\begingroup$ But the value of $Y_i$ is completely determined by not more than the $i$-th toss, right? So if $i\neq j$ then $Y_i$ and $Y_j$ are independent. That is the only independence that I am referring to. I am not saying that the two rv's you mention are independent. If that would be the case then the answer would have been $0$. $\endgroup$
    – drhab
    Dec 24 '20 at 16:11
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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

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  • $\begingroup$ Thanks Henry. However, it is hard to believe Cov(X2, X2+X3) equals to Var(X2) : to get this work, X2 and (X2+X3) need to be uncorrelated, correct ? $\endgroup$
    – user152503
    Dec 24 '20 at 19:15
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    $\begingroup$ @user152503 Not correct. It is enough that $X_2$ and $X_3$ are uncorrelated. On base of that we find: $$\mathsf{Cov}(X_2,X_2+X_3)=\mathsf{Cov}(X_2,X_2)+\mathsf{Cov}(X_2,X_3)=\mathsf{Var}(X_2)+0$$ $\endgroup$
    – drhab
    Dec 24 '20 at 21:29
  • $\begingroup$ Thanks a lot drhab and Henry, now I finally understand the logic! $\endgroup$
    – user152503
    Dec 25 '20 at 8:48
  • $\begingroup$ @user152503 You are very welcome. Also you can accept one of the answers. The one that comes closest to your understanding I would say. Gezegende Kerst toegewenst. $\endgroup$
    – drhab
    Dec 25 '20 at 11:06

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