5.03g Cdf of transformed variables

The continuous random variable \(X\) has probability density function given by $$f(x) = \begin{cases} \frac{1}{2t}x^2 & 1 \leq x \leq 4, \\ 0 & \text{otherwise}. \end{cases}$$ The random variable \(Y\) is defined by \(Y = X^2\). Show that \(Y\) has probability density function given by $$g(y) = \begin{cases} \left(\frac{1}{2t}\right)y^{\frac{1}{2}} & 1 \leq y \leq 16, \\ 0 & \text{otherwise}. \end{cases}$$ [5] Find

1. the median value of \(Y\), [2]
2. the expected value of \(Y\). [2]

The continuous random variable \(X\) has probability density function \(f\) given by $$f(x) = \begin{cases} \frac{1}{80}\left(3\sqrt{x} - \frac{8}{\sqrt{x}}\right) & 4 \leqslant x \leqslant 16, \\ 0 & \text{otherwise}. \end{cases}$$

1. Find the distribution function of \(X\). [3]

The random variable \(Y\) is defined by \(Y = \sqrt{X}\).

1. Find the probability density function of \(Y\). [3]

The continuous random variable \(X\) has distribution function given by $$\text{F}(x) = \begin{cases} 0 & x < 0, \\ \frac{1}{90}(x^2 + x^4) & 0 \leqslant x \leqslant 3, \\ 1 & x > 3. \end{cases}$$ The random variable \(Y\) is defined by \(Y = X^2\).

1. Find the probability density function of \(Y\). [4]
2. Find the mean value of \(Y\). [2]

The random variable \(X\) has probability density function f given by $$\mathrm{f}(x) = \begin{cases} \frac{1}{30}\left(\frac{8}{x^2} + 3x^2 - 14\right) & 2 \leqslant x \leqslant 4, \\ 0 & \text{otherwise}. \end{cases}$$

1. Find the distribution function of \(X\). [3]

The random variable \(Y\) is defined by \(Y = X^2\).

1. Find the probability density function of \(Y\). [4]
2. Find the value of \(y\) such that \(\mathrm{P}(Y < y) = 0.8\). [3]

The continuous random variable \(X\) has probability density function f given by $$f(x) = \begin{cases} \frac{1}{8} & 0 \leq x < 1, \\ \frac{1}{28}(8 - x) & 1 \leq x \leq 8, \\ 0 & \text{otherwise}. \end{cases}$$

1. Find the cumulative distribution function of \(X\). [3]

1. Find the value of the constant \(a\) such that P\((X \leq a) = \frac{5}{7}\). [3]

The random variable \(Y\) is given by \(Y = \sqrt[3]{X}\).

1. Find the probability density function of \(Y\). [5]

The continuous random variable \(X\) has cumulative distribution function given by $$F(x) = \begin{cases} 0 & x < 0, \\ \frac{1}{16}x^2 & 0 \leq x \leq 4, \\ 1 & x > 4. \end{cases}$$

1. The random variable \(Y\) is defined by \(Y = \frac{1}{X^2}\). Find the cumulative distribution function of \(Y\). [5]
2. Show that E\((Y)\) is not defined. [4]