To find the cdf of Y (denoted as FY), we need to evaluate it for different values of t.
For t < 0, Y = FX(X) takes on a value less than 0 with probability 0 (since FX only takes values between 0 and 1). Therefore, FY(t) = 0.
For t >= 1, Y = FX(X) takes on a value greater than or equal to 1 with probability 1 (since FX only takes values between 0 and 1 and maps to [0, 1]). Therefore, FY(t) = 1.
For 0 <= t < 1, we need to find the probability that Y = FX(X) is less than or equal to t. Since FX(X) is the inverse of FX, this is equivalent to finding the probability that X is less than or equal to FX^(-1)(t). We can express this as:
FY(t) = P(Y <= t) = P(FX(X) <= t) = P(X <= FX^(-1)(t)) = FX(FX^(-1)(t)) = t
Therefore, for 0 <= t < 1, FY(t) = t.
Let \, X\, be a random varible with invertible cdf \, F_ X. Define another random variable \, Y=F_ X(X). Find the cdf F_ Y of \, Y\,.
For t<0:
F_ Y(t)=\quad
unanswered
For t\geq 1:
F_ Y(t)=\quad
unanswered
For \, 0\leq t< 1:
F_ Y(t)=\quad
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