# 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:

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 $$corr(X_{2}, X_{2} + X_{3})$$

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 $$\mathsf{Var}Y_1$$.

• 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 Dec 24 '20 at 15:24
• 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 $0$. Dec 24 '20 at 16:11

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

• 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 ? Dec 24 '20 at 19:15
• @user152503 Not correct. It is enough that $X_2$ and $X_3$ are uncorrelated. On base of that we find: $$\mathsf{Cov}(X_2,X_2+X_3)=\mathsf{Cov}(X_2,X_2)+\mathsf{Cov}(X_2,X_3)=\mathsf{Var}(X_2)+0$$ Dec 24 '20 at 21:29
• Thanks a lot drhab and Henry, now I finally understand the logic! Dec 25 '20 at 8:48
• @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. Dec 25 '20 at 11:06