Let $X, Y$ be two independent random variables, $X \sim U[0, 3], Y \sim U[1, 5]$. Calculate $Pr[X < Y − 1]$ and determine distribution of $X + 2Y$ .

Question

I'm on my freshmen year at college and I'm taking a probability course, and I came up with this question and I don't know exacly how to work with that.

I have tried, and that's what I thought.

$X$ is a uniform random variable from $0$ to $3$. Then, the probability $P(X<x) = \int_0^x {1 \over 3} dx$.

And $Y$ is uniform random variable from $1$ to $5$. Then, $P(Y<y) = \int_0^y {1 \over 4} dy$

And now I have to think about a way to compare the the variables. When $X < Y-1$?

Answer 1

Here are geometrical proofs for your two questions. See figures below.

Question 1 : Point $(X,Y)$ being uniformly distributed into rectangle $[0,3] \times [1,5]$, inequation $X<Y-1 \iff Y>X+1$ is realized iff $(X,Y)$ is above diagonal line with equation $y=x+1$, i.e., belongs to the upper trapezoid, whose area represents $5/8$th of the total surface. Otherwise said :

$$P(X<Y-1)=\frac58$$

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2. Question 2 (edit)

Let us work on the cdf of $M=X+2Y$ instead of its pdf.

In the general case :

$$F(m)=P(X+2Y < m) = area(ABNM)/area(ABCD)$$

$$F(m)= \frac{AM+BN}{2} \times AB /(3 \times 4)=\frac{3}{48}(2m-5)$$

Differentiating $F(m)$, we get the density

$$f(m)=\frac{1}{8}$$

i.e., a constant. But this computation is valid for the main cases ($5 \le m \le 10$). There are two particular cases $2<m<5$ (triangle $AN'M'$) and $10<m<13$ (pentagon $ABN''M''D$) that should be treated separately. In fact, we can shortcut these separate treatment ; indeed, as areas' increasing is linear in variable $m$, we just have to make linear connections with the "plateau" of the curve of $f$ : for example, in the first case $2<m<5$ (relative area of triangle $AN'M'$), we construct linear function $f(m)=\frac{1}{24}(m-2)$ which is such that $f(2)=0$ and $f(5)=\frac18$. The final curve of $f$ finaly looks as a isosceles trapezoid.

Answer 2

Once we obtain the joint PDF of continuous random variables $X$ and $Y$ (specified above), say $f_{XY}(x,y)$, we may write any probability of the form $\mathbb{P}(X \leq f(Y))$, as $$ \mathbb{P}(X \leq f(Y)) = \int_{\mathcal{Y}} \int_{0}^{f(Y)} f_{XY}(x,y) \text{d}x \text{d}y $$ where $\mathcal{Y}$ is the appropriate support for random variable $Y$ (I will leave determining this for you).

As for finding the distribution of $Z = aX + bY$, some of the most straightforward methods to find such distribution are though inspection of the moment generating function or characteristic function, or by change of variables again utilizing the joint PDF $f_{XY}(x,y)$. Recall that the MGF for the uniform random variable $U \sim \text{Uniform}[a,b]$ is given by $$ M_U(t) = \frac{e^{tb}-e^{ta}}{t(b-a)} $$

Answer 3

In general, for a subset $A$ of $\Bbb R^2$, $$ \text{Pr}[(X,Y) \in A] = \iint_A f_{X,Y}(x,y) \, dxdy. $$ So, \begin{align} \text{Pr}[X<Y-1] &= \iint_{x<y-1} f_{X,Y}(x,y) \, dxdy \\ &= \int_{-\infty}^\infty \int_{-\infty}^{y-1} f_{X,Y}(x,y) \, dxdy. \end{align} Now recall that if $X$ and $Y$ are independent, $f_{X,Y}(x,y) = f_X(x) f_Y(y)$.

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One way to determine the distribution of $U=X+2Y$ is by using the change of variable theorem:

1. Define $V=X$, and compute the joint density function of $U$ and $V$ with the formula $$ f_{U,V}(u,v) = f_{X,Y}(x,y) \left| \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} \right|, $$ where $x$ and $y$ are written in terms of $u$ and $v$ from the equations $u=x+2y$ and $v=x$.
2. Now compute the marginal density of $U$ and you’re done: $$ f_U(u) = \int_{-\infty}^\infty f_{U,V}(u,v)\,dv. $$