$Suppose that X is a random variable uniform in (0, 1), and define $M = 2 \max\{X, 1− X\} − 1$. Determine the distribution function of M.$

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$Suppose that X is a random variable uniform in (0, 1), and define $M = 2 \max\{X, 1− X\} − 1$. Determine the distribution function of M.$

So far I got to transforming the variable and splitting it because of max into
for x>0.5: (1-x)/2
for x< 0.5: (x-1)/2

Now I should plug it into the CDF for X, but I am unsure what the CDF is with the boundaries given. Any help much appreciated, hope the 5 bucks are enough.

Can anyone explain shortly how to get there and solve the question?
Thanks in advance

Probability

Kilian Schulz

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Let's derive the CDF of $M$ by definition:

Let $t\in [0,1]$, which is the range of $M$: $0\leq M \leq 1$ (notice that $M$ is the largest when $X = 0$ or $X=1$ and $M$ is the lowest when $X= \frac{1}{2}$.

$F_M(t)= P(M\leq t) = P(2\max\{X,1-X\}-1\leq t) = P(\max\{X,1-X\}\leq \frac{t+1}{2}) = P(X \leq \frac{t+1}{2}, 1-X \leq \frac{t+1}{2})$

The last equality is perhaps the most difficult one. But $\max\{X,1-X\}\leq \frac{t+1}{2}$ if and only if $X \leq \frac{t+1}{2}, 1-X \leq \frac{t+1}{2}$.

Continuing,

$F_M(t) = P(X \leq \frac{t+1}{2}, 1-X \leq \frac{t+1}{2}) = P(X \leq \frac{t+1}{2}, 1- \frac{t+1}{2} \leq X) = P(1-\frac{t+1}{2}\leq X \leq \frac{t+1}{2}) = t$, for $t\in[0,1]$

Mathe

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Kilian Schulz

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is P(X<=t) supposed to be t/2 in the equality you called the most difficult one? Otherwise i think im following
- Kilian Schulz
  
  0
  
  sorry this i understand now. but the very last equality is the last thing that confuses me. How does P(1-t/2
Kilian Schulz

0

also i think you forgot the -1 at the end of M
- Mathe
  
  0
  
  I did forget the -1 at the end. Let me fix it!
Mathe

0

I fixed the missing -1 and a smally typo!
Kilian Schulz

0

thanks a lot! youre saving my exam tmrw

$Suppose that X is a random variable uniform in (0, 1), and define $M = 2 \max\{X, 1− X\} − 1$. Determine the distribution function of M.$

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