Continuous Random Variables

Step 1 — write the total-area condition for this support. Since $f(x)=0$ outside $[0,2]$, the whole area is the integral from $0$ to $2$: $\int_0^2 kx^2\,dx = 1$

Step 2 — integrate, treating $k$ as a constant multiplier. $k\left[\frac{x^3}{3}\right]_0^2 = 1$

Step 3 — substitute the limits. $k\left(\frac{8}{3}-0\right) = 1$

Step 4 — solve for $k$. $\frac{8k}{3} = 1 \quad\Longrightarrow\quad k = \frac38$

Step 1 — write the total-area condition. $\int_0^a \frac{kx^2}{a^2}\,dx = 1$

Step 2 — take the constant $\dfrac{k}{a^2}$ outside the integral, then integrate $x^2$. $\frac{k}{a^2}\left[\frac{x^3}{3}\right]_0^a = 1$

Step 3 — substitute the limits. $\frac{k}{a^2}\times\frac{a^3}{3} = 1$

Step 4 — simplify the power of $a$ ($a^3\div a^2=a$), then solve for $k$. $\frac{ka}{3} = 1 \quad\Longrightarrow\quad k = \frac{3}{a}$

Step 1 — recall what has to be shown. "Could be a pdf" means exactly two things, and the mark scheme awards one mark for each: $f(x)\geqslant0$ everywhere, and the total area under the graph is $1$. So the answer needs both, not just the more interesting one.

Step 2 — check non-negativity. The segment runs from $(0,0)$ up to $(2,1)$, so it lies on or above the $x$-axis for every $x$ in $[0,2]$, and $f(x)=0$ outside. Hence $f(x)\geqslant0$ for all $x$ — the first condition holds.

Step 3 — identify the shape under the graph. From $(0,0)$ to $(2,1)$ with the $x$-axis beneath it, the region is a right-angled triangle: base along the $x$-axis of length $2$, and vertical height $1$ at $x=2$.

Step 4 — find the total area with the triangle formula. $\text{Area} = \frac12\times\text{base}\times\text{height} = \frac12\times2\times1 = 1$

Step 5 — state the conclusion. The total area is exactly $1$ and $f(x)\geqslant0$ throughout, so both defining conditions are satisfied and $f$ could be a probability density function.

Step 1 — see which way round the question runs. The previous part checked that the area was $1$. Here we are told $g$ is a pdf, so the area is $1$ — and that single fact is what pins down the one unknown, the radius.

Step 2 — write the area of the region in terms of the unknown. Call the radius $r$. The graph is the upper half of a circle of radius $r$, so the region between it and the $x$-axis is a semicircle, of area $\frac12\pi r^2$.

Step 3 — set that area equal to $1$. $\frac12\pi r^2 = 1$

Step 4 — multiply both sides by $2$, then divide by $\pi$. $\pi r^2 = 2 \quad\Longrightarrow\quad r^2 = \frac{2}{\pi}$

Step 5 — take the square root, keeping the positive root, since a radius cannot be negative. $r = \sqrt{\frac{2}{\pi}} \approx 0.798$

Step 1 — use symmetry about $x=2$ to find the reflection of the given interval. The interval $2$ to $5$ is $3$ units to the right of the line of symmetry; reflecting it across $x=2$ gives the interval $-1$ to $2$, $3$ units to the left. By symmetry, these have equal probability: $P(-1<X<2) = P(2<X<5) = \frac{117}{256}$

Step 2 — use the fact that the two halves of the total area, split at the line of symmetry $x=2$, are each exactly $\dfrac12$. So the "far tail" beyond $-1$ (i.e. $X\leqslant-1$) has probability $\frac12-\frac{117}{256} = \frac{128}{256}-\frac{117}{256} = \frac{11}{256}$

Step 3 — assemble $P(X>-1)$ as the complement of that far tail. $P(X>-1) = 1-\frac{11}{256} = \frac{245}{256}$

Step 1 — find the new support, mapping each endpoint of $X$'s support through $y=2x-1$. $x=0 \;\mapsto\; y=2(0)-1=-1, \qquad x=4 \;\mapsto\; y=2(4)-1=7$ So $Y$ takes values on $[-1,7]$.

Step 2 — find the new width, and compare with the old. Old width $=4-0=4$; new width $=7-(-1)=8$ — exactly double, matching the multiplier $a=2$.

Step 3 — find the new height, using the fact that width $\times$ height must still equal $1$. $\text{new height} \times 8 = 1 \quad\Longrightarrow\quad \text{new height} = \frac18 = 0.125$

Step 4 — write the density of $Y$ explicitly. $g(y) = 0.125, \qquad -1\leqslant y\leqslant7$

Step 5 — describe the graph. Compared with $f$, $g$'s graph is the same flat shape, shifted and stretched: twice as wide, positioned from $-1$ to $7$ instead of $0$ to $4$, and exactly half as tall.

Step 1 — read off what you need about $X$ itself. The graph is symmetric about $x=1$, so $E(X)=1$. Its support is $[-1,3]$, so its width is $3-(-1)=4$, and its greatest height is $0.375$. Those three numbers — centre $1$, width $4$, height $0.375$ — are all that either sketch depends on.

Step 2 — (a) find $S$'s centre. $E(S)=2\times E(X) = 2\times1 = 2$, so $S$'s graph is centred on $s=2$ instead of $x=1$ — shifted one unit to the right.

Step 3 — (a) find $S$'s width. $Var(S)=Var(X)$ — the variance is unchanged, so the spread is unchanged and the width is still $4$. A width of $4$ centred on $2$ runs from $2-2=0$ to $2+2=4$.

Step 4 — (a) find $S$'s height. The width did not change, so the height must not change either, or the area would stop being $1$. The peak stays at height $0.375$, now sitting above $s=2$.

Step 5 — (a) describe the sketch. The same quadratic shape, from $s=0$ to $s=4$, touching the axis at both ends, with its maximum at the point $(2,\,0.375)$.

Step 6 — (b) find $T$'s centre. $E(T)=E(X)=1$, so this graph does not move — it stays centred on $1$.

Step 7 — (b) find $T$'s width. $Var(T)=\frac14Var(X)$. Width follows the standard deviation, not the variance, so take the square root: $\sqrt{\tfrac14} = \tfrac12$ The width is halved, from $4$ to $2$. A width of $2$ centred on $1$ runs from $1-1=0$ to $1+1=2$.

Step 8 — (b) find $T$'s height. The width halved, so the height must double to keep the area at $1$: $0.375\times2 = 0.75$

Step 9 — (b) describe the sketch. The same quadratic shape but narrower and taller, from $t=0$ to $t=2$, with its maximum at the point $(1,\,0.75)$.