## I. INTRODUCTION Rukhin(1988) introduced a loss function given by, $$ L (\theta , \delta , \gamma) = w (\theta , \delta) \gamma^ {- \frac {1}{2}} + \gamma^ {\frac {1}{2}} \tag {1.1} $$ Where, $\gamma$ is an estimator of the loss function $w(\theta,\delta)$, which is non-negative. Guobing(2016) used this loss function and derived estimates of the loss and risk function of the parameter of Maxwell's distribution. Singh (2021) took various forms of $w(\theta,\delta)$ and derived estimates of the loss and risk function of the parameter of a continuous distribution which gives Half-normal distribution, Rayleigh distribution and Maxwell's distribution as particular cases. Rukhin(1988) considered the Bayesian estimation of the unknown parameter $\theta$ of the binomial distribution by taking $$ w (\theta , \delta) = (\theta - \delta) ^ {2} \tag {1.2} $$ In this paper, Bayes estimate of the unknown parameter $\theta$ of the binomial distribution has been obtained by replacing $w(\theta, \delta)$ by $w_1(\theta, \delta)$ given by $$ w _ {1} (\theta , \delta) = h (\theta) (\theta - \delta) ^ {2} \tag {1.3} $$ Where, $$ h (\theta) = \frac {1}{\{\theta (1 - \theta) \}} \tag {1.4} $$ ## II. ESTIMATION OF LOSS AND RISK OF THE PARAMETER OF BINOMIAL DISTRIBUTION Let the random variable $X$ follows binomial distribution with parameters $n$ and $\theta$. Where $\theta$ is unknown satisfying $0 \leq \theta \leq 1$. The prior p.d.f of $\theta$, denoted by $\pi_1(\theta)$ is as follows: $$ \pi_{1} (\theta) = \left\{ \begin{array}{l l} \frac{\theta^{\alpha - 1} (1 - \theta)^{\beta - 1}}{B (\alpha , \beta)} & \text{if } \alpha \geq 0, \beta \geq 0, 0 < \theta < 1 \\ 0 & \text{Otherwise} \end{array} \right. \tag{2.1} $$ Under the assumption of prior probability density function (p.d.f.) for $\theta$ as above, Bayes estimates of $\theta$ derived by Rukhin (1988) were as follows: For $\alpha \geq 0, \beta \geq 0$ $$ \delta_ {B} (X) = \frac {(X + \alpha)}{(n + \alpha + \beta)} \tag {2.2} $$ $$ \gamma_ {B} (X) = \frac {(X + \alpha) (n + \beta - X)}{(n + \alpha + \beta) ^ {2} (n + \alpha + \beta + 1)} \tag {2.3} $$ and for $\alpha = 0, \beta = 0$ $$ \delta_ {0} (X) = \frac {X}{n} \tag {2.4} $$ $$ \gamma_ {0} (X) = \frac {X (n - X)}{n ^ {2} (n + 1)} \tag {2.5} $$ It was shown that $$ E _ {\theta} L \left(\theta , \delta_ {0}, \gamma_ {0}\right) = \infty \tag {2.6} $$ Under, $w_{1}(\theta,\delta)$ as above, the corresponding Bayes estimate is given by, For $\alpha \geq 0,\beta \geq 0$ $$ \delta_ {1 B} (X) = \frac {E \left\{\theta h (\theta) / X \right\}}{E \left\{h (\theta) / X \right\}} \tag {2.7} $$ $$ \delta_ {1 B} (X) = \frac {(X + \alpha - 1)}{A - 2} \tag {2.8} $$ On simplification, provided, $A = n + \alpha +\beta >2$ and, $$ \gamma_ {1 B} (X) = E \left\{\theta h (\theta) / X \right\} - \left\{\delta_ {1 B} (X) \right\} ^ {2} E \left\{h (\theta) / X \right\} \tag {2.9} $$ $$ \gamma_ {1 B} (X) = \frac {1}{A - 2} \tag {2.10} $$ on simplification, provided, $A = n + \alpha +\beta >2$ We, see that, in this case $\gamma_{1B}(X)$ does not depend upon $X$ and is function of $n,\alpha$ and $\beta$ $$ E_{\theta} L\left(\theta,\delta_{1B},\gamma_{1B}\right) = E_{\theta}\left[h(\theta)\left(\theta - (X + \alpha - 1)(A - 2)^{-1}\right)^{2}\right](A - 2)^{1/2} + (A - 2)^{-1/2} $$ Or, $$ E_{\theta} L\left(\theta,\delta_{1B},\gamma_{1B}\right) = \left[n+h(\theta)\left(1-\alpha+\theta(\alpha+\beta-2)\right)^{2}\right]\left(A-2\right)^{-3/2} + \left(A-2\right)^{-1/2} < \infty \tag{2.12} $$ In this case, $$ R (\theta , \delta_ {1 B}) = E _ {\theta} \{h (\theta) (\theta - \delta_ {1 B}) \} ^ {2} \tag {2.13} $$ Or, $$ R (\theta , \delta_ {1 B}) = [ n + h (\theta) \{1 - \alpha + \theta (\alpha + \beta - 2) \} ^ {2} ] (A - 2) ^ {- 2} \tag {2.14} $$ As mentioned by Keifer (1977), an estimator $\gamma(X)$ is said to be conservatively biased if, $$ E _ {\theta} \left\{\gamma (X) \right\} \geq R (\theta , \delta) = E _ {\theta} \left\{w (\theta , \delta) \right\} \tag {2.15} $$ In the light of this condition, $\gamma_0(X)$ as given by Rukhin (1988) is not conservatively biased. In this case, $$ E _ {\theta} \left\{\gamma_ {1 B} (X) \right\} = \frac {1}{A - 2} \tag {2.16} $$ Let $\delta_{0B}(X)$ and $\gamma_{0B}(X)$ be values of $\delta_{1B}(X)$ and $\gamma_{1B}(X)$, respectively when, $\alpha = \beta = 0$. If possible let, $$ E _ {\theta} \left\{\gamma_ {0 B} (X) \right\} \geq R (\theta , \delta_ {0 B}) \tag {2.17} $$ which holds if, $$ - 2 \theta^ {2} + 2 \theta - 1 \geq 0 \tag {2.18} $$ which is a contradiction, since $0 < \theta < 1$ and maximum value of $-2\theta^2 + 2\theta - 1$ is $-\frac{1}{2}$ which corresponds to $\theta = \frac{1}{2}$. Moreover, $-2\theta^2 + 2\theta - 1 = -1$ for $\theta = 1$ and $\theta = 0$. Thus, $\gamma_{0B}(X)$ is not conservatively biased. When $\alpha = \beta = 1$, we have, $$ E _ {\theta} \left\{\gamma_ {1 B} (X) \right\} = R (\theta , \delta_ {1 B}) = \frac {1}{n} \tag {2.19} $$ When, $\alpha = \beta > 1$, $\theta = 0.5$ $$ E _ {\theta} \left\{\gamma_ {1 B} (X) \right\} \geq R (\theta , \delta_ {1 B}) \tag {2.20} $$ When, $\alpha = \beta > 1$, $\theta \neq 0.5$ $$ E _ {\theta} \left\{\gamma_ {1 B} (X) \right\} \geq R (\theta , \delta_ {1 B}) \tag {2.21} $$ which holds if $$ \alpha \leq 1 + g (\theta) \tag {2.22} $$ .Where, $$ g (\theta) = \frac {2 \theta (1 - \theta)}{(2 \theta - 1) ^ {2}} \tag {2.23} $$ $g(\theta)$ is a monotonically increasing function of $\theta$ over the set $S = (0,1) - \{0.5\}$. Hence, $\gamma_{1B}(X)$ as above, presents a valid 'frequentist report' as mentioned by Berger(1985). The results are summarized in the following: THEOREM. Let $(\delta_{1B},\gamma_{1B})$ be Bayes estimators of the unknown parameter $\theta$ of the binomial distribution under the loss function $L(\theta,\delta,\gamma) = \frac{1}{\{\theta(1 - \theta)\}} (\theta -\delta)^2\gamma^{-\frac{1}{2}} + \gamma^{\frac{1}{2}}$ and beta prior density with known parameters $\alpha$ and $\beta$. Then, the frequentist risk $E_{\theta}L(\theta,\delta_{1B},\gamma_{1B})$ is finite for all values of $\alpha$ and $\beta$ provided $0 < \theta < 1$. For $\alpha = \beta = 0$, $\gamma_{1B}(X)$ is not conservatively biased. The estimator $\gamma_{1B}(X)$ is conservatively biased for $\alpha = \beta = 1$ and for $\alpha = \beta >1$ satisfying $\alpha \leq 1 + \frac{2\theta(1 - \theta)}{(2\theta - 1)^2},\theta \neq 0.5$. However, if $\alpha = \beta >1,\theta = 0.5$, $\gamma_{1B}(X)$ is also conservatively biased.

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