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Suppose we have a number of independent random variables of the form $X_1 \sim U[a_1,b_1], X_2 \sim U[a_2,b_2], X_3 \sim U[a_3,b_3]$. Now, suppose we generate a random variable $Y$ as follows:

$$Y = \max_{1 \leq l \leq n} \left(1 - \left(\sum_{i=0}^{l-1} \binom{n}{i} X_1^i \cdot (1-X_1)^{n-i} \right) \cdot (1-X_2^3 \cdot X_3)\right) \cdot \frac{l}{n}$$

I want to find $\mathbb{E}[Y]$. Does it make sense to approach this approach with Monte-Carlo simulation?

Thanks in advance.

Regards.

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  • $\begingroup$ Are you sure that all $X_i$ are not standard uniforms? That would make so much more sense. $\endgroup$ – A.S. Mar 18 '16 at 22:50
  • $\begingroup$ Actually, the $X_i$ are uncertain probabilities, or probability estimators?! Not sure about the wording. I think of something of the form $X_1 \sim U[0.1,0.2]$. So, we are not sure about the probability of a certain event to happen, but we estimate the probability will be something between 0.1 and 0.2, uniformly distributed. $\endgroup$ – Frank Merzen Mar 18 '16 at 22:56
  • $\begingroup$ Then it's clear that the $\max$ is achieved at $l=n$. The most-inner sum is $1-X_1^n$, hence $Y=1-(X_1^n)(1-X_2^3X_3)$ and you can easily compute its expectation by using independence. $\endgroup$ – A.S. Mar 18 '16 at 23:04
  • $\begingroup$ Excuse me, there was a mistake in my post. Could you have a look at it now again? But my question is of a more general nature. Imagine, I would like to identify the max for some $x \leq l \leq y$. So, in my case, I really identify different values of $l$ where the max is achieved (through simulation). So, could you imagine any situation here where MC simulation is needed? Or can I obtain the expected value of such formulas always without simulation? Thanks. $\endgroup$ – Frank Merzen Mar 18 '16 at 23:13
  • $\begingroup$ You have so many parameter choices, even if all of the uniforms are $Unif(0,1),$ that this would turn into a massive simulation project. However, simulation for a few special cases might give some guidance toward an analytical solution. $\endgroup$ – BruceET Mar 19 '16 at 1:31

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