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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

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  • $\begingroup$ 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
    Commented Jan 23, 2017 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
    Commented Jan 24, 2017 at 1:34
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    $\begingroup$ @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
    Commented Jan 24, 2017 at 5:56

1 Answer 1

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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.

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