0
$\begingroup$

$X$ and $Y$ are independent variables that follow the exponential distribution with the same parameter $\beta$. What is the probability $$P(X \geq k\cdot Y)$$ if $k>0$ ?

I don't really know where to start.

$\endgroup$
  • $\begingroup$ Set up and evaluate a double integral. $\endgroup$ – André Nicolas Jun 9 '16 at 13:38
2
$\begingroup$

$X$ and $Y$ follow an exponential distribution. Therefore, the density function is $f(t)= \beta \exp{(-\beta t)}$. First, we condition by $Y=k$ and integrates over the $k$.

$$P(X\geqslant kY) = \int_0^\infty \beta \exp{(-\beta t)} P(X\geqslant kt| Y=t)dt$$

Since $X$ and $Y$ are independent, $P(X\geqslant kt | Y=t)=P(X\geqslant kt)=\exp(-\beta \times kt)$. Plugging it in the previous equality: $$P(X\geqslant kY) = \int_0^\infty \beta \exp{(-\beta (k+1)t)}dt$$ Which by integration gives: $$\frac{\beta}{\beta (k+1)}=\frac{1}{k+1}$$

$\endgroup$
1
$\begingroup$

Required probability $$ \iint_{x\ge ky\ge0}p_{X,Y}(x,y)dA=\int_{0}^\infty\int_{ky}^\infty p_{X,Y}(x,y)dxdy\\ =\int_{0}^\infty\int_{ky}^\infty p_X(x)p_Y(y)dxdy=\beta^2\int_{0}^\infty\int_{ky}^\infty e^{-\beta(x+y)}dxdy\\ =\beta\int_0^{\infty}-e^{-\beta(x+y)}\Big{|}_{ky}^\infty dy=\beta\int_0^{\infty}e^{-\beta(ky+y)} dy={1\over k+1} $$

$\endgroup$
0
$\begingroup$

Hint: For fixed $k>0$, $$\Pr(X\geq kY)=\int_0^{\infty}\Pr(X\geq ky)f_Y(y)\;dy$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.