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Let $X$ be a continuous random variable with probability density function $f_X$ given by

$f_X(x) =$

$x + 3$ for $− 3 ≤ x ≤ −2$
$3 − x$ for $2 ≤ x ≤ 3$
$0$ otherwise.

Find q0.1 and q0.9.


I'm unsure as to how to use the probability density function to find the quantiles? Could someone guide me in which direction to go?

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  • $\begingroup$ Do you know how to get the cumulative distribution function (CDF) from the probability density function (pdf)? $\endgroup$ – hardmath Jul 6 '14 at 17:52
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The $0.1$ quantile or $10^o$ percentile can be obtained by identifying the value $j$ on the x-axis where integrating $ (x+3) dx$ between $-3$ and $j$ we get $1/10$.

The indefinite integral is $x^2/2+3x+c$. Thus we have to find $j$ so that $j^2/2+3j-9/2 +9$ equals $1/10$. This leads to $$j^2/2+3j+22/5=0$$

$$j^2+6j+44/5=0$$

from which $ j=-3+\frac{1}{\sqrt{5}}$.

The $0.9$ quantile or $90^o$ percentile can be obtained in a similar manner by identifying the value $k$ on the x-axis where integrating $(3-x) dx$ between $k$ and $3$ we get $1/10$.

The indefinite integral is $3x-x^2/2+c$. As above, we have to find $k$ so that $9-9/2-3k+k^2/2$ equals $1/10$. This yields $$k^2/2-3k+22/5=0$$

$$k^2-6k+44/5=0$$

from which $ j=3-\frac{1}{\sqrt{5}}$.

Lastly, note that for this problem we can follow a shorter way to the solution. If you graph the density function, it simply includes two triangular portions, the first with vertices in $(-3,0)$, $(-2,0)$, $(-2,1)$, and the second with vertices in $(3,0)$, $(2,0)$, $(2,1)$. The two portions are equal and symmetric right triangles with both legs equal to $1$. The area of each portion is therefore $1/2$. Considering by simplicity only the left triangular portion, we must find a value on the x-axis $j>-3$ so that the area of the right triangle with vertices in $(-3,0)$, $(j,0)$, $(j,j+3)$ (which represents a leftward small portion of the original pdf) is equal to $1/10$. The length of both legs of this small triangle are given by the distance $d$ between $-3$ and $j$, so that we have $d^2/2=1/10$, which directly leads to $d=1/\sqrt{5}$. The value of $x$ identifying the 0.1 quantile is therefore $ j=-3+d=-3+\frac{1}{\sqrt{5}}$.

Thanks to the symmetry of the graph, similar considerations applied to the right triangular portion allows to get that the the $ 0.9$ quantile is identified by $k=3-\frac{1}{\sqrt{5}}$.

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  • $\begingroup$ Was it all ok with your quantile calculations? Had you other problems? $\endgroup$ – Anatoly Jul 10 '14 at 10:37
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Quantile gives you the value $q$, for which $P(X \leq x) = q$. So in the first case you need to solve $\int_{-\infty}^{a}f(x)dx = q$ for a. Also, in the problem you gave the probability is negative.

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  • $\begingroup$ I think those may be dashes $\endgroup$ – Kamster Jul 6 '14 at 16:48
  • $\begingroup$ Yeah, sorry I edited the format to make it correct. Would I still go from negative infinity? $\endgroup$ – Eric Jul 6 '14 at 16:58
  • $\begingroup$ It goes from your lower bound ($-3$) $\endgroup$ – Alex Jul 6 '14 at 17:40
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for F(x)=0,1 you have to calculate $\int_{-2}^x (t+3) \quad dt=0.1$ Now find x.

greetings, calculus

Edit: With you last correction the condition $\int_{-\infty}^{\infty} f(x)\ \ dx=1$ is fullfilled.

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