1
$\begingroup$

This question is based on this one . Let $X=(X_1,X_2)$ be a bivariate random variable with the joint PDF $$f(x_1,x_2)=\frac{1}{2}I_{[0,4]}x_1 I_{[0,\frac{1}{4}x_1]}x_2$$

First : I've asked whether this function a valid PDF is ?. The answer was that if I integrate $x_2$ form $0$ up to$\frac{x_1}{4}$ I can easily and clearly verify that's a valid PDF , but now my new question(s) is , is the upper bound or the maximum value of $x_2$ equal to $1$ or not ? (since the maximum value $x_1$ can take is 4 ) , if the answer is yes , why can not we integrate $x_2$ up to $1$ which in in turn proves that is not a valid PDF ! $$\int_0^4 \int_0^1 0.5 \, dx_2 \, dx_1 = 2 \tag 1$$.

Second :I want to find $P(2\leq X_1\leq 3;\frac{1}{2}\leq X_2 \leq \frac{3}{2})$ using the joint CDF i.e $$P(2\leq X_1\leq 3;\frac{1}{2}\leq X_2 \leq \frac{3}{2}) = F(3;\frac{3}{2}) -F(3;\frac{1}{2})-F(3;\frac{3}{2})+F(2:\frac{2}{2})$$

I have tried to find the CDF but unfortunately it seems that I miss a very basic idea either in probability theory or in calculus (or both ). That's my attempt .

By definition $F(b_1,b_2)=P(X_1\leq b_1,X_2 \leq b_2)$

If $b_1 < 0 $ and/or $b_2 < 0,$ then $$F(b_1,b_2)=0$$ If $b_1 \in [0,4]$ and $b_2 \in [0,\frac{b_1}{4}]$ then $$F(b_1,b_2)=\int_0^{b_1} \int_0^{b_2} \frac{1}{2} \, dx_2 \, dx_1=\frac{1}{2}b_1b_2\tag 2$$ If $b_1 \in [0,4]$ and $b_2 > \frac{b_1}{4}$ then $$F(b_1,b_2)=\frac{1}{16}b_1^{2}$$ If $b_1 > 4 $ and $b_2 \in [0,\frac{b_1}{4}]$ then $$F(b_1,b_2)=2b_2$$ Otherwise $F(b_1,b_2)=1$ .

There is absolutely something wrong in deriving this CDF , suppose I have to find $P(x_1 \leq 3,x_2 \leq 0.6)$ using the this derived CDF I will get $3.6 > 1$ !.

Please can someone help me by answering my questions or explaining where I am wrong . Thanks a lot in advance

$\endgroup$
  • $\begingroup$ For the first question, i.e. validity of the pdf, you can not integrate over [0,1] for $x_2$ since for a specific $x_1$, $x_2$ should be in the given range: e.g. if you integrate over [0,1] you are implicitly assuming that for $x_1=1$, $x_2$ can also be 1, but it cannot! $\endgroup$ – tempx Jan 23 '17 at 6:45
  • $\begingroup$ @tempx please can you elaborate this , if $x_1 = 1$ $x_2$ can also be $1$ but it cannot ! I mean if $x_1=1$ $x_2$ can not exceed $\frac{1}{4}$ $\endgroup$ – Geo Aßaad Jan 24 '17 at 1:34
  • 1
    $\begingroup$ @geo-abaad I mean for the pdf calculation if you take the integral limits as [0,4] and [0,1] , itmeans that your are assuming that $x_2$ is in [0,1] independent of $x_1$: e.g. for $x_1<1$, $x_2$ can be any number in [0,1] which is outside of the defined range. ($x_2$ should be $x_2<1/4$) $\endgroup$ – tempx Jan 24 '17 at 5:56
1
$\begingroup$

The integration $(1)$ is wrong. If we integrate the common density with respect to $x_1$ at a given $0\leq x_2\leq 1$ then (as shown below)

enter image description here

the correct integral is

$$\frac12\int_{4x_2}^4\ dx_1=2(1-x_2),$$

which is the density belonging to the random variable $X_2$. Indeed, this is a density:

$$2\int_0^1 (1-x_2)\ dx_2=2(1-\frac12)=1.$$

As far as the cdf of $X_1,X_2$: Take a look at the following figure in which five different cases are shown:

enter image description here

The common cdf, the probability of the event $\{X_1<x_1\cap X_2<x_2\}$ can be calculated by integrating the common pdf over the blue regions above. For example in the first two cases the results are $1$ and $0$, respectively.

$\endgroup$

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.