# Joint cumulative distribution of dependent variables

#### OhMyMarkov

##### Member
Hello everyone!

The problem:
$X,Y,Z$ are random variables that are dependent and uniformly-distributed in $[0,1]$, and let $\alpha$ be a given number in $[0,1]$. I am asked to compute the following:

$\text{Pr}(X+Y+Z>\alpha \;\;\; \& \;\;\; X+Y\leq \alpha)$​

What I have so far

$f_{X+Y+Z,X+Y}(u,v)=f_{Z,X+Y}(u-v,v)=f_{Z}(u-v)\cdot f_{X+Y}(v)$

(1) Is the above equation correct? I think it stands for discrete RVs but not quite sure for continuous RVs... If it is true, is the following integral correct to compute the desired probability?

$\int _{\alpha} ^{+\infty} \int _{-\infty} ^{\alpha} f_{X+Y+Z,X+Y}(u,v) du\; dv = \int _{\alpha} ^{+\infty} \int _{-\infty} ^{\alpha} f_{Z}(u-v)f_{X+Y}(v) du\; dv= \int _{\alpha} ^{+\infty} \alpha f_{X+Y}(v) dv$​

#### girdav

##### Member
I've not checked carefully the details but it follows from a substitution. I think we can simplify the last integral.

#### OhMyMarkov

##### Member #### chisigma

##### Well-known member
Hello everyone!

The problem:
$X,Y,Z$ are random variables that are dependent and uniformly-distributed in $[0,1]$, and let $\alpha$ be a given number in $[0,1]$. I am asked to compute the following:

$\text{Pr}(X+Y+Z>\alpha \;\;\; \& \;\;\; X+Y\leq \alpha)$​

What I have so far

$f_{X+Y+Z,X+Y}(u,v)=f_{Z,X+Y}(u-v,v)=f_{Z}(u-v)\cdot f_{X+Y}(v)$

(1) Is the above equation correct? I think it stands for discrete RVs but not quite sure for continuous RVs... If it is true, is the following integral correct to compute the desired probability?

$\int _{\alpha} ^{+\infty} \int _{-\infty} ^{\alpha} f_{X+Y+Z,X+Y}(u,v) du\; dv = \int _{\alpha} ^{+\infty} \int _{-\infty} ^{\alpha} f_{Z}(u-v)f_{X+Y}(v) du\; dv= \int _{\alpha} ^{+\infty} \alpha f_{X+Y}(v) dv$​
If X, Y and Z are non negative r.v., then if X + Y + Z > a then X + Y < a so that what You have to find is the p.d.f. of the r.v. T = X + Y + Z. X, Y and Z are uniformely distributed in [0,1], so that their p.d.f. and its L-transform is...

$\displaystyle f(t) = \mathcal {U} (t) - \mathcal{U} (t-1) \implies F(s) = \frac{1 - e^{-s}}{s}\ (1)$

The p.d.f. of T is the convolution of three p.d.f. ...

$\displaystyle f_{T} (t) = f(t)*f(t)*f(t) = \mathcal{L}^{-1}\{\frac{1 - 3\ e^{- s} + 3\ e^{- 2\ s} - e^{- 3\ s}}{s^{3}}\} = \frac{1}{6}\ \{ t^{2}\ \mathcal U(t) - 3\ (t-1)^{2}\ \mathcal U (t-1) + 3\ (t-2)^{2}\ \mathcal {U} (t-2) - (t-3)^{2}\ \mathcal{U} (t-3)\}\ (2)$

Now is simply...

$\displaystyle P \{ X + Y + Z > a\} = 1 - \int_{0}^{a} f_{T}(t)\ dt\ (3)$

Kind regards

$\chi$ $\sigma$

#### girdav

##### Member There is just a minor typo in the initial post (you meant the random variables are independent). Now I think it's good. Did you compute a density of $X+Y$.

#### Dhamnekar Winod

##### Active member
If X, Y and Z are non negative r.v., then if X + Y + Z > a then X + Y < a so that what You have to find is the p.d.f. of the r.v. T = X + Y + Z. X, Y and Z are uniformely distributed in [0,1], so that their p.d.f. and its L-transform is...

$\displaystyle f(t) = \mathcal {U} (t) - \mathcal{U} (t-1) \implies F(s) = \frac{1 - e^{-s}}{s}\ (1)$

The p.d.f. of T is the convolution of three p.d.f. ...

$\displaystyle f_{T} (t) = f(t)*f(t)*f(t) = \mathcal{L}^{-1}\{\frac{1 - 3\ e^{- s} + 3\ e^{- 2\ s} - e^{- 3\ s}}{s^{3}}\} = \frac{1}{6}\ \{ t^{2}\ \mathcal U(t) - 3\ (t-1)^{2}\ \mathcal U (t-1) + 3\ (t-2)^{2}\ \mathcal {U} (t-2) - (t-3)^{2}\ \mathcal{U} (t-3)\}\ (2)$

Now is simply...

$\displaystyle P \{ X + Y + Z > a\} = 1 - \int_{0}^{a} f_{T}(t)\ dt\ (3)$

Kind regards

$\chi$ $\sigma$
Hello,
What is $\mathcal U(t)-\mathcal U(t-1)$? Is it a Probability density function of uniform distribution at t=1. Have you used here heavyside function concept? Why did you multiply (2) by $\frac16$ ? It should be multiplied by $\frac12$. My guess is X,Y,Z are dependent random variables, therefore, you multiplied (2) by $\frac16 i.e. (\frac12*\frac13)$ Am i correct?

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