3
$\begingroup$

Let $X$ be a stochastic variable $X \sim R(0,1)$ - Thus $F_X(x) = 1$ if $x \in ]0,1[$ otherwise $0$

Let $Z$ be a stochastic variable, independent of $X$, $Z \sim b(1, 1/2)$.

$$P(Z = 0) = P(Z = 1) = 1/2$$

Let $U = X + Z$

Find distribution function of $U$. $F_U(u) = P(U \leq u) = P(X + Z \leq u)$

Prove that the distribution is $R(0,2)$

I have been trying to solve this problem for hours now, but can write the solution down. Intuitively $U$ will be distribued $R(0,2)$ since every real number $]0,1[$ $[1,2[$ has equal change by the fact $Z \sim b(1, 1/2)$

Please give me some advice and sketch out how to solve the problem.

Best regards Nicolas

$\endgroup$
6
  • $\begingroup$ Please fix your formulas using LaTeX, e.g. $X\sim R(0,1)$ gives you $X\sim R(0,1)$. What is the distribution class $R$? $\endgroup$
    – SBF
    Jan 15, 2013 at 12:11
  • $\begingroup$ Thanks, never tried LaTeX but I will look into it. The distribution class $R(0,1)$ it the real numbers $]0,1[$ It just means that every real number in this interval is mapped to $density = 1$ using the mass-density function $f_X(x)$ $\endgroup$
    – Shuzheng
    Jan 15, 2013 at 12:14
  • $\begingroup$ The uniform distribution, you mean? $\endgroup$
    – SBF
    Jan 15, 2013 at 12:15
  • $\begingroup$ Exactly, sorry im not very specific - this is my first course in probability as a "freshmen" $\endgroup$
    – Shuzheng
    Jan 15, 2013 at 12:16
  • 1
    $\begingroup$ Your use of notation is ambiguous. If $F$ denotes the cumulative probability density function (CDF) of a random variable, since you say $F_U(u) = P(U \leq u)$, then $F_X(x) = 1$ for $0 < x < 1$ is incorrect; for a random variable uniformly distributed on $]0,1[$, $F_X(x) = P(X\leq x)$ equals $x$, not $1$ for $x \in ]0,1[$. $\endgroup$
    – Dilip Sarwate
    Jan 15, 2013 at 12:33

1 Answer 1

Reset to default
1
$\begingroup$

Hint: prove that $$\forall u \in [0,1],\quad P(U \leq u) = P(X \leq u,\; Z = 0)$$ and $$\forall u \in [0,1],\quad P(1 <U \leq 1 + u) = P(X \leq u,\; Z = 1)$$

$\endgroup$
10
  • $\begingroup$ But $u \in [0,2]$ right ? $\endgroup$
    – Shuzheng
    Jan 15, 2013 at 12:19
  • $\begingroup$ As you mentioned in the OP, it is convenient to split up the the interval. $\endgroup$
    – Siméon
    Jan 15, 2013 at 12:23
  • $\begingroup$ $P(X<=u, Z=0)$ is equal to the integral with upper bound $u$ and lower bound $0$ of $fX(x) * 1/2$ due to independence right ? And $P(X<=u, Z=1)$ will always be $1/2$ ? If $Z = 1$ then $X < u$ since $u = X + Z$ and $X \in ]0,1[$ $\endgroup$
    – Shuzheng
    Jan 15, 2013 at 12:29
  • $\begingroup$ I am not sure what you mean. Because of independence, $P(X \leq u, Z = 0) = P(X \leq u)P(Z=0) = \int_0^udx \times \frac{1}{2} = ...$ $\endgroup$
    – Siméon
    Jan 15, 2013 at 12:34
  • $\begingroup$ I think I get it, although i think splitting the interval up make it more confusing, since $u \in ]0,2[$ I appreciate your help guys. $\endgroup$
    – Shuzheng
    Jan 15, 2013 at 12:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .