Question

A point is chosen at random inside a triangle with vertices at (0, 0), (0, 8), and (8, 0). The continuous random variable ???? denotes the x-coordinate of that point.

a) Find the probability that ???? is less than 5.

b) Find the cumulative distribution function (i.e. P(???? ≤ x) ).

c) Find the probability density function.

d) Find the average value of X.

Answer 1

Assuming all points in the triangle
`T` are uniformly distributed, we have the joint density

`f_(X,Y)(x,y)=\begin{cases}\frac1A&\text{for }(x,y)\in T\\0&\text{otherwise}\end{cases}`

where
`A=\frac{8^2}2=32` is the area of the triangle
`T`.

(a)

`P(X<5)=\displaystyle\iint_(T^*)f_(X,Y)(x,y)\,\mathrm dy\,\mathrm dx`

(where
`T^*` is the portion of
`T` for which
`x` is between 0 and 5)

`P(X<5)=\displaystyle\frac1{32}\int_0^5\int_0^(8-x)\mathrm dy\,\mathrm dx`

`P(X<5)=\displaystyle\frac1{32}\int_0^5(8-x)\,\mathrm dx`

`P(X<5)=\frac1{32}\cdot\frac{55}2=\boxed{(55)/(64)}`

(b) Generalizing the previous result, we have

`P(X\le x^*)=\displaystyle\iint_(T^*)f_(X,Y)(x,y)\,\mathrm dy\,\mathrm dx`

(this time with
`T^*` being the portion of
`T` where
`0\le x\le x^*` for some
`x^*` between 0 and 8)

`P(X\le x^*)=\displaystyle\frac1{32}\int_0^(x^*)\int_0^(8-x)\mathrm dy\,\mathrm dx`

`P(X\le x^*)=\displaystyle\frac1{32}\int_0^(x^*)(8-x)\,\mathrm dx`

`P(X\le x^*)=\displaystyle\frac1{32}\left(8x^*-\frac{(x^*)^2}2\right)`

That is, the CDF of
`X` is

`P(X\le x)=\begin{cases}\frac{8x-\frac{x^2}2}{32}&\text{for }0\le x\le8\\0&\text{otherwise}\end{cases}`

or

`\boxed{P(X\le x)=\begin{cases}(16x-x^2)/(64)&\text{for }0\le x\le8\\0&\text{otherwise}\end{cases}}`

(c) Obtain the PDF by differentiating the CDF:

`f_X(x)=(\mathrm d)/(\mathrm dx)P(X\le x)`

`\boxed{f_X(x)=\begin{cases}(8-x)/(32)&\text{for }0<x<8\\0&\text{otherwise}\end{cases}}`

(d) Compute the expectation of
`X`:

`E[X]=\displaystyle\int_0^8xf_X(x)\,\mathrm dx`

`E[X]=\displaystyle\frac1{32}\int_0^8x(8-x)\,\mathrm dx=\boxed{\frac83}`