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Let $X_j$ be a random variable that is $1$ with probability $x_j^*$, and $0$ with probability $1-x_j^*$. The random variables $X_j$ are independent and $j$ belongs to $\{1,\ldots,n\}$ for some positive integer $n$. I would like to calculate the mean value of $\max\limits_{j}X_j$. That is,

$$\mathbb{E}\left[\max\limits_{j}X_j\right]$$

My try is: since $X_j$ is a binary random variable, than the maximum would be $1$. Therefore, $$\mathbb{E}\left[\max\limits_{j}X_j\right]=\Pr\left[X_j=1\right]=x_j^*,$$ but I do not know if this is the right argument.

Kaalouss
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3 Answers3

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It means $\max \{X_j : 1 \leq j \leq n\}$.

Your reasoning actually doesn't even make sense. When you write $\mathbb E[\max_j X_j] = \mathbb{Pr}[X_j=1]$, the reader absolutely wonders what on earth $j$ refers to in the second expression. You turned a free variable to a bound variable by sleight of hand.

The idea, which you should formalize: If each $X_j \in \{0,1\}$, then $\max = 1$ if any are 1 and 0 otherwise. So calculate the probability that any equal 1.

djechlin
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The max is either $0$ or $1$.

The probability that it's $0$ is the probability that all (independent) $X_j$ are $0$, that is

$$P(max X_j =0)=\prod_j P(X_j=0)=\prod_j (1-x^*_j)$$

And of course, $$P(max X_j =1)=1-\prod_j (1-x^*_j)$$

Thus $E(max X_j)=1-\prod_j (1-x^*_j)$

Jean-Claude Arbaut
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$\max_j X_j=0$ if all the variables $X_i$'s are $=0$ (which happens with probability $\prod_{j=1}^n (1-x_j^\star)$) and $=1$ if at least one of the $X_j$'s is equal to $1$ (which happens with probability $1-\prod_{j=1}^n (1-x_j^\star)$ by normalization). Therefore the mean value of $\max_j X_j$ is $$ \mathbb{E}[\max_j X_j]=1\times \left[1-\prod_{j=1}^n (1-x_j^\star)\right]+0\times \prod_{j=1}^n (1-x_j^\star)=1-\prod_{j=1}^n (1-x_j^\star)\ . $$

Pierpaolo Vivo
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