Expectation integration with conditions

To calculate the conditional expectation \( E[Y_1 | Y_1 + Y_2 + \cdots + Y_n = y] \), we start by recognizing that \( Y_1, Y_2, \ldots, Y_n \) are independent and identically distributed (iid) nonnegative random variables. We denote \( S_n = Y_1 + Y_2 + \cdots + Y_n \). Given the symmetry and identical distribution of the \( Y_i \)'s, we can intuitively understand that the conditional expectation \( E[Y_1 | S_n = y] \) should be the same for each \( Y_i \). Therefore, we have: \[ E[Y_1 | S_n = y] = E[Y_2 | S_n = y] = \cdots = E[Y_n | S_n = y]. \] Since there are \( n \) such random variables, we can express the conditional expectation in terms of the total sum: \[ E[Y_1 | S_n = y] = \frac{1}{n} E[S_n | S_n = y]. \] Now, \( E[S_n | S_n = y] = y \) because if we condition on \( S_n \) being equal to \( y \), then \( S_n \) is exactly \( y \). Putting this together, we find: \[ E[Y_1 | S_n = y] = \frac{1}{n} y. \] Thus, the final result for the conditional expectation is: \[ \boxed{\frac{y}{n}}. \]

To compute the unconditioned expectation \( E[Y_1] \), we can use the following integral expression: \[ E[Y_1] = \int y_1 p(y_1, y_2, \dots, y_n) \, dy_1 \, dy_2 \, \cdots \, dy_n, \] where \( p(y_1, y_2, \ldots, y_n) \) represents the joint probability density function (PDF) of the random variables \( (Y_1, Y_2, \ldots, Y_n) \).

Given that \(Y_1, Y_2, \ldots, Y_n\) are independent and identically distributed (i.i.d.), we can express the expected value \(E[Y_1]\) through the following integration: \[ E[Y_1] = \int y_1 \prod_{i=1}^n f(y_i) \, dy_1 \, dy_2 \, \ldots \, dy_n = \int y_1 f(y_1) \, dy_1. \] The transition to the second equality is justified by the normalization property of the probability density function.

In analyzing $E[Y_1|Y_1+\cdots+Y_n=y]$, we can express it as follows: \begin{equation} E[Y_1|Y_1+\cdots+Y_n=y]=\int{y_1 \cdot p(y_1, y_2, \ldots, y_n | Y_1 + Y_2 + \cdots + Y_n = y) \, dy_1 \, dy_2 \ldots dy_n} \, , \end{equation} where $p(y_1, y_2, \ldots, y_n | Y_1 + Y_2 + \cdots + Y_n = y)$ denotes the condi