Probability that order statistic is larger than the other

Question

Given the density function:

$$f_Y(y)=e^{-(y+1)}, y>-1$$

Let $Y_1,..,Y_4$ be a random sample from the distribution defined by the density function above. Let $Y_{(1)},..,Y_{(4)}$ be the order statistics of $Y_1,..,Y_4$.

Find $P(Y_{(1)}<3Y_{(2)})$

Edit 3: Seems I have it right with finding the joint distribution for this. However, according to the formula:

$$f_{Y(i),Y(j)}(y_i,y_j)=\frac{n!}{(i-1)!(j-i-1)!(n-j)!}*F(y_i)^{i-1}*[F(y_j)-F(y_i)]^{j-i-1}*[1-F(y_j)]^{n-j}*f(y_i)*f(y_j)$$

I get:

$$f_{Y(1),Y(2)}(y_1,y_2)=12*e^{-3(y_2+1)}*e^{-(y_1+1)}, y_1,y_2>-1$$

Which is not a probability function. What am I doing wrong?

Furthermore, I think I should do the completing probability as it is easier, and simply have 1-* to reach the probability I seek. Do I have the double integral right at least:

$$\int_{-1}^0\int_{-1}^\frac{Y_1}{3}f_{Y(1),Y(2)}(y_1,y_2)dy_2dy_1$$

I believe so and the result I get is probably off because of having some mistake with the joint distribution function, which I simply can not spot.

Answer 1

The joint distribution of $(Y_{(1)},Y_{(2)})$ is actually $$f_{Y_{(1)},Y_{(2)}}(y_1,y_2)=12\,\mathrm e^{-3(y_2+1)}\,\mathrm e^{-(y_1+1)}\,\mathbf 1_{y_2\gt y_1>-1}, $$ hence $$ P(Y_{(1)}\lt3Y_{(2)})=\iint f_{Y_{(1)},Y_{(2)}}(y_1,y_2)\,\mathbf 1_{y_1\lt3y_2}\,\mathrm dy_1\mathrm dy_2. $$ It happens that somewhat simpler computations are involved if one considers $$ P(Y_{(1)}\gt3Y_{(2)})=\iint f_{Y_{(1)},Y_{(2)}}(y_1,y_2)\,\mathbf 1_{y_1\gt3y_2}\,\mathrm dy_1\mathrm dy_2, $$ since $y_1\gt3y_2$ and $y_2\gt y_1>-1$ imply $y_1\lt0$ hence $$ P(Y_{(1)}\gt3Y_{(2)})=\int_{-1}^0\int_{y_1}^{y_1/3}12\,\mathrm e^{-3(y_2+1)}\,\mathrm e^{-(y_1+1)}\,\mathrm dy_2\mathrm dy_1=\int_0^14\mathrm e^{-x_1}\int_{x_1}^{(x_1+2)/3}3\mathrm e^{-3x_2}\,\mathrm dx_2\mathrm dx_1, $$ and maybe you can finish?

Answer 2

This is mechanized in Maple. Note that Maple uses the standard definition of the order statistics. $$restart; with(Statistics): X := RandomVariable(Distribution(PDF = (y ->piecewise(-1 < y, exp(-y-1), 0) ))): Y1 := OrderStatistic(X, 1, 4): PDF(Y1, t); $$ The latter command outputs $$\cases{0&$t\leq -1$\cr 4\,{{\rm e}^{-4-4\,t}}&$-1<t$\cr}$$ Next $$Y2 := OrderStatistic(X, 2, 4): PDF(Y2, t); $$ produces $$ \cases{0&$t\leq -1$\cr 12\,{{\rm e}^{-2-2\,t}} \left( {{\rm e}^{-1-t}}-1 \right) ^{2}&$-1<t$\cr}$$ At last, $$Probability(Y1 < 3*Y2); $$ gives $$-9/4\,{{\rm e}^{-8/3}}+{\frac {16\,{{\rm e}^{-2}}}{5}}=0.2767351410. $$ See Distribution and OrderStatistic for info.