0

I encountered an expected value of a max function issue where: $$ E[\max(\frac{vX+1}{2}, \frac{1}{2})] $$ $v$ follows a uniform distribution on $[0,1]$ and $X$ follows a binomial distribution. I know that I should separate the max function and take integrals, but not sure how to deal with $vX$ there. Note that $v$ and $X$ are independent.

cpiegore
  • 1,556
lemon
  • 9
  • 2
    Are $v$ and $X$ independent? – MPW Aug 11 '23 at 18:21
  • Yes, I will edit the question to include it! – lemon Aug 11 '23 at 18:26
  • You just need to condition i.e. $$E[f(v,X)] = E[E[f(v,X)\mid X] = E[E[f(v,X)\mid v]]$$ Since $X,v$ are independent, you have, for example, $$E[f(v,X)\mid X] = \varphi(X),\quad \varphi: x\mapsto E[f(v,x)]$$ – Andrew Aug 11 '23 at 18:43

1 Answers1

0

To my mind, for the equation $ E[\max(\frac{vX+1}{2}, \frac{1}{2})] $ the "max" part of it means:

$\max(\frac{vX+1}{2}, \frac{1}{2}) = \begin{cases} \frac{vX+1}{2} &\text{if}\ \frac{vX+1}{2} > \frac{1}{2}, \\ \frac{1}{2} &\text{otherwise}.\end{cases}$

but if I divide out $\frac{vX+1}{2} = \frac{vX}{2} + \frac{1}{2}$ and since $v, X \geq 0$, then the question boils down to finding the expected value of just $\frac{1}{2} + \frac{vX}{2}$.

Since $v ,X$ are independent, then $E[\frac{1}{2} + \frac{vX}{2}] = \frac{1}{2} + \frac{1}{2}\cdot E[v] \cdot E[X]$. For the expected value of a uniform distribution $E[v] = 0.5$ since it is a special case of $E[max(Y)]$ for some $n$ draws from a uniform distribution $Y$ for which the formula is $\frac{n}{n+1}$, in this case $n = 1$, so $E[v] = 0.5$. For the expected value of binomial distribution $E[X] = np$, where $n$ is the number of draws and $p$ is the probability of success. So combining everything:

$E[\max(\frac{vX+1}{2}, \frac{1}{2})] = E[\frac{vX+1}{2}] =\frac{1}{2} + \frac{1}{2} \cdot E[v] \cdot E[X] = \frac{np + 2}{4}$

Gustamons
  • 81