S3 2012 6 OCR

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(i)

We should have:

\displaystyle \int_{-\infty}^\infty f(t) \mathrm dt = 1

that is:

\displaystyle a \int_0^1 t \mathrm dt + a \int_1^4 \mathrm dt = 1

so:

\displaystyle \frac 1 2 a + 3a = 1

giving:

\displaystyle \frac 7 2 a = 1

so:

\displaystyle a = \frac 2 7

(ii)

For t < 0, we have f(t) = 0, so \mathrm P(T \le t) = 0 for t < 0.

For 0 \le t \le 1 we have:

\displaystyle \mathrm P(T \le t) = \frac 2 7 \int_0^t x \mathrm dx = \frac {t^2} 7

For 1 < t \le 4 we have:

\begin{align*}\mathrm P(T \le t) & = \frac 2 7 \int_0^1 x \mathrm dx + \frac 2 7 \int_1^t \mathrm dx \\ & = \frac 1 7 + \frac 2 7 \left(t - 1\right) \\ & = \frac 2 7 t - \frac 1 7\end{align*}

since \mathrm P(T \le 4) = 1, we have \mathrm P(T \le t) = 1 for t > 4, giving the CDF of T as:

\displaystyle \mathrm P(T \le t) = \begin{cases}0 & t < 0\\ \frac {t^2} 7 & 0 \le t \le 1 \\ \frac 2 7 t - \frac 1 7 & 1 < t \le 4 \\ 1 & t \ge 4\end{cases}

(iii)

The CDF of Y is given by:

\displaystyle F_Y(y) = \mathrm P(T^{1/2} \le y) = \mathrm P(T \le y^2)

where y \ge 0. Substituting this into the above we have:

\begin{align*}\mathrm P(T \le y^2) &= F_T(y^2) \\ & = \begin{cases}0 & y^2 < 0\\ \frac {(y^2)^2} 7 & 0 \le y^2 \le 1 \\ \frac 2 7 y^2 - \frac 1 7 & 1 < y^2 \le 4 \\ 1 & y^2 \ge 4\end{cases} \\ & = \begin{cases}0 & y < 0\\ \frac {y^4} 7 & 0 \le y \le 1 \\ \frac 2 7 y^2 - \frac 1 7 & 1 < y \le 2 \\ 1 & y \ge 2\end{cases}\end{align*}

in the latter equality we’ve just rewritten the equalities in y^2 in terms of y noting that y \ge 0.

Differentiating with respect to y, we find the PDF of Y to be:

\displaystyle f_Y(y) = \begin{cases}\frac 4 7 y^3 & 0 \le y \le 1 \\ \frac 4 7 y & 1 < y \le 2 \\ 0 & \text {otherwise}\end{cases}

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