91Ó°ÊÓ

The cumulative distribution function (CDF) is the integral of the probability density function (PDF). In this case, we are given the PDF of \(X\): \[f(x) = \begin{cases} 1-e^{-x} & \text{ for } x\geq 0 \\ 0 & \text{ for } x < 0 \end{cases}\] The CDF, denoted by \(F(x)\), is defined as: \[F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t) dt\]

We will break up the integral into two parts: from \(-\infty\) to 0 and from 0 to \(x\). \[F(x) = \int_{-\infty}^{x} f(t) dt = \int_{-\infty}^{0} f(t) dt + \int_{0}^{x} f(t) dt\] Since \(f(t) = 0\) for \(t < 0\), the first integral is just 0. \[F(x) = 0 + \int_{0}^{x} f(t) dt\] Now, we will calculate the second integral: \[\int_{0}^{x} f(t) dt = \int_{0}^{x} (1-e^{-t}) dt\] To integrate this function, utilize the linearity of the integral for both terms separately: \[\int_{0}^{x}(1-e^{-t})dt = \int_{0}^{x} dt - \int_{0}^{x} e^{-t} dt\] Integrate each part, and we get: \[\left[t - [-e^{-t}]\right]_{0}^{x} = \left[x - (-e^{-x})\right] - \left[0 - (-e^{0})\right] = x - (-e^{-x}) - e^{0}\] So, the CDF of \(X\) is given by: \[F(x) = \begin{cases} 0 & \text{for } x < 0 \\ x - (-e^{-x}) - e^{0} & \text{for } x \geq 0 \end{cases}\]

Now we need to find the transformation \(Y = \sqrt{X}\). Let \(G(y)\) be the distribution function for \(Y\). We know that: \[G(y) = P(Y \leq y) = P(\sqrt{X} \leq y) = P(X \leq y^2) = F(y^2)\] So we need to substitute \(y^2\) into the CDF: \[G(y) = \begin{cases} 0 & \text{for } y^2 < 0 \\ y^2 - (-e^{-y^2}) - e^{0} & \text{for } y^2 \geq 0 \end{cases}\] The square of \(y\) can never be negative, so the first case is not applicable. Hence, the distribution function of \(Y\) is given by the second case: \[G(y) = y^2 - (-e^{-y^2}) - e^{0} = y^2 + e^{-y^2} - 1\] Therefore, the distribution function \(G(y)\) for \(Y = \sqrt{X}\) is: \[G(y) = y^2 + e^{-y^2} - 1\]