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June 27, 2018
Xiao-Li Meng writes: BFF again? Yes, and this time it is for Best Friends Forever, not another Bayesian, Fiducial and Frequentist workshop – for that you need to wait until BFF6. [For BFF# history, see https://imstat.org/2017/05/xl-files-bayesian-fiducial-and-frequentist-bff4ever/.] But what on Earth is BGF? Best Goal Forever (FIFA time)? Best…

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George Roussas writes in memory of his colleague: Joining in the celebration of the life and achievements of PK Bhattacharya [see his obituary], I wish to scribe a few words based on reminiscences from our lives on campus of UC-Davis. PK was present almost from the establishment of Statistics…

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Prodyot Kumar Bhattacharya passed away March 9, 2018, at his home in Davis, California. He was professor emeritus at the University of California at Davis and contributed to the field of statistics during a career that spanned more than 50 years. PK, as he was called by many colleagues and…

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Dr. George Edward Cave of West New York, son of the late George Thomas Cave and the late Leatha Willa Cave, née Shelton, passed away at age 71 on April 18, 2018 in Washington DC. While attending Memorial High School, George was President of the General Student Organization, a National…

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Congratulations to Mirza Uzair Baig at the University of Hawai’i at Mānoa, who wrote an excellent solution to the problem. Note that the statistic Tn may be represented as \[ T_n = I_{Y_{(1)} < X_{(1)}, Y_{(n)} < X_{(n)}}\, \bigg [\sum_{i = 1}^n I_{Y_i < X_{(1)}} + \sum_{i = 1}^n I_{X_i > Y_{(n)}}\bigg ]\] \[+ \,I_{X_{(1)} < Y_{(1)}, X_{(n)} < Y_{(n)}}\, \bigg [\sum_{ i = 1}^n I_{X_i < Y_{(1)}} + \sum_{i = 1}^n I_{Y_i > X_{(n)}}\bigg ].\] Denote the empirical CDF of $X_1, \cdots , X_n$ by $F_n$…

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