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Let $x$ 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 here 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 ! $$\int_0^4 \int_0^1 0.5 \, dx_2 \, dx_1 = 2 $$.

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 $$F(b_1,b_2)=\int_0^{b_1} \int_0^{b_2} \frac{1}{2} \, dx_2 \, dx_1=\frac{1}{2}b_1b_2$$ 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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This is obviously a self-study problem, and the first thing you should do is to edit your question to include the self-study tag.

The second thing you should do is to draw a sketch of the plane with coordinate axes $x_1$ and $x_2$ and indicate on it which region of the plane it is on which the joint pdf has value $\frac 12$. Then set up the double integral that you need to compute, taking particular care about the limits. Don't wing it; look at the diagram very very carefully and pay attention to the fact that you have a nested integral in which, as far as the inner integral with respect to $x_2$ is concerned, $x_1$ is just some fixed number in $(0,4)$, say $\pi$ if you have difficulty thinking of one. For this value of $x_1$, for what values of $x_2$ is the joint pdf $f(\pi,x_2)$ nonzero? Plug in the formula and don't forget to replace all occurrences of $x_1$ by $\pi$, on the left side as well as the right side.

Actually, if you have any memory of what you were taught in calculus about what an integral is calculating and what a double integral is calculating, you might be able to deduce the value of the double integral from a formula that you most likely had to memorize in middle school but since then have forgotten.

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