Uniform distribution in (0,1). P(X1+X2<=X3) and Gaussian RV with variance 1/4 and 1/9 , P(3V>=2U)

Question

I'm appearing for a competitive examination and I find a lot of questions from probability involving $2$ or more random variables are very common. Please help me with the method on how to deal with these questions when you have comparisons of RV(ie <>= inside the bracket). Listing a few of them :

1. $U$ and $V$ be two independent zero mean Gaussian RV, with variances $1/4$ and $1/9$. Then $P(3V \geq 2U) ?$
2. $X_1, X_2$ and $X_3$ are independent RV's with Uniform distribution in $(0,1)$. $P(X_1+X_2 \leq X_3)$

For the first one I tried as follows, $$W=3V-2U ; P(W \geq 0); \operatorname{Mean}(W)=0; \operatorname{Variance}(W)= (9 \cdot 1/9)+(4 \cdot 1/4) =2.$$ Now I don't know how to proceed from here. One more doubt I have here, Is $W$ a Gaussian RV?

For the second problem I know that $X_1+X_2$ will have a PDF that is a triangle for $0$ to $2$ with max value $1$. But I don't know how to account for the comparison. Please help.

Answer 1

Hints:

• If $U$ and $V$ are independent and normally distributed then any linear combination $W=aU+bV$ is normally distributed.

• $$P\left\{ X_{1}+X_{2}\leq X_{3}\right\} =\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}f\left(x,y,z\right)dxdydz$$ where $f$ is prescribed by $\left(x,y,z\right)\mapsto1$ if $x+y\leq z$ and $\left(x,y,z\right)\mapsto0$ otherwise.

Answer 2

For the probability that $X_1+X_2\le X_3$, we can use your observation about the distribution of $X_1+X_2$. Call this random variable $X$, and call $X_3$ by the name $Y$.

The random variable $X$ has density function $f(x)$ which is $x$ on $(0,1]$ and $1-x$ on $(1,2)$.

The density function of $Y$ is $1$ on $(0,1)$, so by independence the joint density of $X$ and $Y$ is $f(x)$ on the rectangle whose corners are $(0,0)$, $(2,0)$, $(2,1)$, and $(0,1)$, and $0$ elsewhere.

Draw the rectangle. We want the probability that $X\le Y$. Draw the line $y=x$. We want the probability that $(X,Y)$ lies above the line, but of course in the rectangle, so in a certain triangle. That triangle has corners $(0,0)$, $(1,1)$, and $(0,1)$. In that triangle, the joint density function of $X$ and $Y$ is $x$. So we actually only need the first half of the sawtooth part.

Thus we want $\iint_T x\,dA$ where $T$ is the triangle. This double integral can be expressed as the iterated integral $$\int_{y=0}^1\left(\int_{x=0}^y x\,dy\right)\,dx.$$ The calculation is easy. Alternately we can integrate first with respect to $y$, with $y$ going from $x$ to $1$.

Remark: The triple integral approach of drhab is better, and generalizes nicely. If we want the probability that $X_1+X_2+X_3\le X_4$, we certainly do not want to first find the full distribution of $X_1+X_2+X_3$.