Any point in the interval [0, 1) can be represented by its decimal expansion 0.x1x2 . . .. suppose a point is chosen at random from the interval [0, 1). let x be the first digit in the decimal expansion representing the point. compute the density of x.
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Note that every point x ∈ [0, 1) has a unique decimal expansion except the rational numbers : x = 0.x1x2 . . . xN = 0.x1x2 . . . xN−1(xN − 1)99 . . .
where x1, x2, . . . , xN ∈ {0, 1, . . . , 9} with xN ≥ 1.
Hence the only points which do not have a unique first digit in its decimal expansion (considering the above ambiguous cases) are 1/10, 2/10,. . . , 9/10 corresponding to N = 1, as above.
For these numbers we would disallow the second kind of expansion, which mean the allowed decimal expansion of i/10 is
i/10 = 0.i (and not 0.(i − 1)999 . . .) for i = 1, 2, . . . , 9.
With this note, the random variable X (which take values 0, 1, . . . , 9) is well defined. Then X(ω) = i if ω ∈ [ i 10 , i+1 10 ) and hence we can assign the probability P(X = i) = 1/10 which is the length of the interval [ i/10, i+1/10 ).
The density function f : R → R is thus defined as
f(x) = ( 1/10 if x = 0, 1, . . . , 9,
0 ; otherwise
which come from the relation f(x) = P(X = x)
where x1, x2, . . . , xN ∈ {0, 1, . . . , 9} with xN ≥ 1.
Hence the only points which do not have a unique first digit in its decimal expansion (considering the above ambiguous cases) are 1/10, 2/10,. . . , 9/10 corresponding to N = 1, as above.
For these numbers we would disallow the second kind of expansion, which mean the allowed decimal expansion of i/10 is
i/10 = 0.i (and not 0.(i − 1)999 . . .) for i = 1, 2, . . . , 9.
With this note, the random variable X (which take values 0, 1, . . . , 9) is well defined. Then X(ω) = i if ω ∈ [ i 10 , i+1 10 ) and hence we can assign the probability P(X = i) = 1/10 which is the length of the interval [ i/10, i+1/10 ).
The density function f : R → R is thus defined as
f(x) = ( 1/10 if x = 0, 1, . . . , 9,
0 ; otherwise
which come from the relation f(x) = P(X = x)
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