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Given process is, Y_{t}-Y_{t-1}=e_{t}-0.5 e_{t-1} The backward shift operator B when applied shifts the data one period back and if we apply B for n number of times,then the data is shifted n period back.  The backward shift operator is such that,  B Y_{t}=Y_{t-1} and in general B^{n} Y_{t}=Y_{t-n} Thus we can write  Y_{t}-Y_{t-1}=e_{t}-0.5 e_{t-1}  as   Y_{t}-B Y_{t}=e_{t}-0.5 B e_{t} (1-B) Y_{t}=(1-0.5 B) e_{t} Thus, the correct option is (a) Y_{t}-Y_{t-1}=e_{t}-0.5 e_{t-1} can be written as (1-B) Y_{t}=(1-0.5 B) e_{t} using backward shift operator. Option b) is incorrect since,  B Y_{t}=(1-0.5 B) e_{t} is simplified as, Y_{t-1}=e_{t}-0.5 B e_{t} Y_{t-1}=e_{t}-0.5 e_{t-1} which is not same as in given question. Option c) is incorrect since, (1-B) Y_{t}=0.5 B e_{t} can be simplified as, Y_{t}-B Y_{t}=0.5 B e_{t} Y_{t}-Y_{t-1}=0.5 e_{t-1} which is not same as in given question. Option d) is incorrect since option a) is correct.   ...