1)p=1/8
2)二项分布
3)EY = np = 5/8,DY = np(1-p) = 5*1/8*7/8 =35/64
EY^2 = DY+(EY)^2 = 35/64 + 25/64 =60/64 = 15/16
追问![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/c9fcc3cec3fdfc030f7af43bd93f8794a4c22636?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
这个能救救吗
哭了 感谢!!!
追答EX = (1+θ)/2 = ∑Xi/n
所以矩估计为θhat = 2*∑Xi/n - 1
L(x) = (1/1-θ)^n
lnL(x) = -n*ln(1-θ)
dlnL(x) /dx = n/(1-θ) >=0
所以lnL单调递增。故极大似然估计 θhat = min(Xi)
追问谢谢了
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