Determine the upper and lower bounds of the expected value of a random variable

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If you increase some of the probabilities of success, the expected number of trials will decrease; if you decrease some of them, it will increase. Thus the case where all probabilities are $\beta$ and the one where they're all $\alpha$ provide bounds on the expected number of trials.

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Milos
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Milos

Updated on March 03, 2020

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  • Milos
    Milos over 3 years

    Let $p_k$ represent the probability of success in the $k$th trial ($k=1,2,\ldots$). It is only known that the following holds $$\alpha \leq p_k \leq \beta$$. There is no other information about $p_k$.

    I want to determine the bounds for the expected number of trials before the first success. Let's denote it as $E(K)$. This is what I've derived:

    \begin{eqnarray*}E(K) &=& \sum_{k=1}^{\infty}\left(kp_k\prod_{j=1}^{k-1}(1-p_j)\right)\\ E(K) &\leq& \sum_{k=1}^{\infty}k\beta(1-\alpha)^{k-1} = \frac{\beta}{\alpha^2}\\ E(K) &\geq& \sum_{k=1}^{\infty}k\alpha(1-\beta)^{k-1} = \frac{\alpha}{\beta^2} \end{eqnarray*}

    However, in the textbook, the solution to this problem is:

    $$\frac 1 \beta \leq E(K) \leq \frac 1 \alpha$$

    It seems that the author just replaced $p_k$ with $\alpha$ and $\beta$:

    $$ k\beta(1-\beta)^{k-1} \leq kp_k\prod_{j=1}^{k-1}(1-p_j)\leq k\alpha(1-\alpha)^{k-1}$$

    My problem with this is that we have $p_k \geq \alpha$ and $\prod_{j=1}^{k-1}(1-p_j) \leq (1-\alpha)^{k-1}$. If we multiply the left sides, which inequality sign should be chosen to relate it with the product of the right sides? I believe that, starting from some $k$, it will really hold that $kp_k\prod_{j=1}^{k-1}(1-p_j)\leq k\alpha(1-\alpha)^{k-1}$, but it is not the case for every $k$. The similar holds for the lower bound.

    So, my questions are as follows:

    • Are the bounds $\frac 1 \beta \leq E(K) \leq \frac 1 \alpha$ correct?

    • If not, are the bounds $\frac \alpha {\beta^2} \leq E(K) \leq \frac \beta {\alpha^2}$ the tightest we can derive?