Skip to main content
added what I think is the calculation for $E[X_1\cdot X_2]
Source Link
EliT
  • 51
  • 1
  • 1
  • 7

Update to reflect what I think is the calculation for $E[X_1\cdot X_2]$

Given n people, if I want to estimate how many of them are likely to have an overlapping birthday with any other person, how do I calculate the variance? So far I have $E[X]=n\cdot p$ with $n$ as the number of people and $p$ the probability of any single person having an overlapping birthday

I think $E[X_1\cdot X_2]$ is as follows:

Using total probability+the fact that $X_i$ is an indicator variable: $E[X_1\cdot X_2]=P_{X_1,X_2}(1)=P(X_1\cdot X_2 |X_1=1)+P(X_1\cdot X_2 |X_1=0)$

Now again as $X_i$ is an indicator variable:

$P(X_1\cdot X_2 |X_1=0)=0$

Now from Bayes conditioning:

$P(X_1\cdot X_2 |X_1=1)=P(X_1 $and$ X_2)\cdot P(X_2)$

This is: $P(X_1\cdot X_2 |X_1=1)\cdot (1-(\frac{364}{365})^n)$

What I'm not sure about is whether: $P(X_1 $and$ X_2)=(1-(\frac{363}{364})^{n-3}+ \frac{1}{365})$

If I'm right:$P(X_1\cdot X_2 |X_1=1)\cdot (1-(\frac{364}{365})^n)=(1-(\frac{363}{364})^{n-3}+ \frac{1}{365})\cdot (1-(\frac{364}{365})^n)=E[X_1\cdot X_2]$

Given n people, if I want to estimate how many of them are likely to have an overlapping birthday with any other person, how do I calculate the variance? So far I have $E[X]=n\cdot p$ with $n$ as the number of people and $p$ the probability of any single person having an overlapping birthday

Update to reflect what I think is the calculation for $E[X_1\cdot X_2]$

Given n people, if I want to estimate how many of them are likely to have an overlapping birthday with any other person, how do I calculate the variance? So far I have $E[X]=n\cdot p$ with $n$ as the number of people and $p$ the probability of any single person having an overlapping birthday

I think $E[X_1\cdot X_2]$ is as follows:

Using total probability+the fact that $X_i$ is an indicator variable: $E[X_1\cdot X_2]=P_{X_1,X_2}(1)=P(X_1\cdot X_2 |X_1=1)+P(X_1\cdot X_2 |X_1=0)$

Now again as $X_i$ is an indicator variable:

$P(X_1\cdot X_2 |X_1=0)=0$

Now from Bayes conditioning:

$P(X_1\cdot X_2 |X_1=1)=P(X_1 $and$ X_2)\cdot P(X_2)$

This is: $P(X_1\cdot X_2 |X_1=1)\cdot (1-(\frac{364}{365})^n)$

What I'm not sure about is whether: $P(X_1 $and$ X_2)=(1-(\frac{363}{364})^{n-3}+ \frac{1}{365})$

If I'm right:$P(X_1\cdot X_2 |X_1=1)\cdot (1-(\frac{364}{365})^n)=(1-(\frac{363}{364})^{n-3}+ \frac{1}{365})\cdot (1-(\frac{364}{365})^n)=E[X_1\cdot X_2]$

edited tags
Link
EliT
  • 51
  • 1
  • 1
  • 7
Source Link
EliT
  • 51
  • 1
  • 1
  • 7

Figuring out the variance of the birthday paradox

Given n people, if I want to estimate how many of them are likely to have an overlapping birthday with any other person, how do I calculate the variance? So far I have $E[X]=n\cdot p$ with $n$ as the number of people and $p$ the probability of any single person having an overlapping birthday