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Hey I am trying to calculate,

\begin{align*} \textbf{E}\left[\max(X_1,X_2,X_3)\right] \end{align*}

Where $X_i$ ase iid Bernoulli distributions with probability $p$. I really just do not know how to approach calculating this. I feel like it should be simple.

Tsangares
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  • If $X=\max{X_1,X_2,X_3}$ do you know how to find the pmf of $X$, namely $p_X$? – Matthew H. Dec 11 '20 at 01:09
  • @MatthewPilling Not really. Is it just the max of the individual Bernoulli pmf? – Tsangares Dec 11 '20 at 01:13
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    No. You need to find $p_X(0)$ and $p_X(1)$. Once you find these values, your expectation becomes $$E(X)=0\cdot p_X(0) + 1\cdot p_X(1)$$ – Matthew H. Dec 11 '20 at 01:14
  • @MatthewPilling $P_X(0) = (1-p)^3$ and $P_X(1) = p^3$ – Tsangares Dec 11 '20 at 01:18
  • @MatthewPilling Okay so $E(X)=E(max(X_1,X_2,X_3) = p^3$? – Tsangares Dec 11 '20 at 01:19
  • @MatthewPilling I see an error I made, $P_X(0) = (1-p)^3$, $P_X(1)=p$ (because only one need to succeed). Making $E(X)=p$ – Tsangares Dec 11 '20 at 01:21
  • @MatthewPilling or is $P_X(1) = (3p + 2p^2 + p^3)/6$? – Tsangares Dec 11 '20 at 01:29
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    $p_X(0)=(1-p)^3$ is correct, which means that $$p_X(1)=1-p_X(0)=1-(1-p)^3$$ For ${X=1}$ to occur we need at least one of the $X_i$'s to be 1. The probability of such an event occuring is not equal to $p$. To find this probability directly, you need to examine in how many ways we can get at least one $X_i$ to equal $1$ – Matthew H. Dec 11 '20 at 01:39

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