Though this question is quite basic, I'm having a hard time understanding it. This is the question from my textbook:

"A random variable $X$ is uniformly distributed in the interval $(-k, k)$. Find $k$ if $P(X \ge 1) = \frac 13 $"

My working:

From what we know about uniform probability distributions, the probability mass function $f(x)$ would be $\frac {1}{2k} $ (This is correct since the area of the rectangle formed in the given interval is $=1$).

I thought this question has something to do with cumulative distribution function. And so I found the CDF (which I found by integrating $\int_{-k}^x f(y) dy $ ), which turned out to be:

CDF $= \begin{cases} 0 & x \gt k \\ \frac x{2k}\ + \frac 12 & -k\le x \lt k \\ 0 & x \ge k\end{cases} $

Since $P(X \ge 1) = \frac 13 $, we could find $P( X \le 1 ) = F(x) = \frac 23$

I tried equating $F(1)$ with $\frac 23$, which gave me $k = -3$. The negative value of $k$ doesn't seem right to me. When I omit the negative sign, I seem to get the correct value.

Is this correct? Or is there a proper way of approach?

  • 2
    $\begingroup$ But $F(1) = \frac{1}{2k} + \frac{1}{2} = \frac{2}{3} \implies \frac{1}{2k} = \frac{1}{6} \implies k=3$. $\endgroup$ – Shirish Kulhari May 8 '18 at 10:39

Well, the CDF is $\mathsf P(X\leqslant x)=F_X(x)=\begin{cases}0 &:& x< -k\\ (x+k)/2k &:& -k\leqslant x< k\\ 1 &:& k\leqslant x\end{cases}$

(Recall that a CDF must have $\lim_{x\to\infty} F_X(x)=1$ among other things.)

So, if we assume $1\leqslant k$, then we have $\mathsf P(X\geqslant 1)=(k-1)/2k$ which is to equal $1/3$

Therefore we find that $\dfrac{k-1}{2k}=\dfrac 13\iff 3(k-1)=2k\iff k-3=0\iff\boxed{k=3}$

[Test: $(3-1)/6=1/3 ~~\checkmark$]

So it seems you just made a sligh tmistake in the alebraic rearangement.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.