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Let $X$ be a random varaible from a distribution with pdf $$ f(x) = \theta x^{\theta-1}, \quad 0< x < 1. $$

a) Name the distribution of $U=-\ln(X)$ by first finding its density

b) Let $X_1, X_2, \ldots,X_n$ be independent and identically distributed random variables with pdf given by earlier with $\theta$= 3. Using the result from a) and by the central limit theorem (CLT)

i) find an approximation to $P(X_1 \cdot X_2 \cdot \ldots\cdot X_{30} \leq 1.85 \cdot 10^{-5})$

i.e. for the probability of the product of the r.v's.

My attempt

I found part a) by doing the transformation, and got an exponential with parameter $\theta$ where my pdf is: $$ f(x)=\theta e^{-u\theta} . $$ Now where I am struggling is with part i) where I am supposed to find an approximation using the CLT and together with the mean equal to $1/3$ and variance equal to $1/9$. I am able to show that with 30 observation we just take the product of every mean and variance from each individual observation giving us:

mean = $(\frac{1}{3})^{30}$
variance = $(\frac{1}{9})^{30}$

and then we can just substitute these into the z score by using central limit theorem $$ P(X_1 \cdot X_2 \cdot \ldots \cdot X_{30} \leq 1.85 \cdot 10^{-5}) = P\left(\frac{X -\mu }{\sigma}\leq \frac {1.85 \cdot 10^{-5} - (\frac{1}{3})^{30}}{(\frac{1}{9})^{15}} \right) $$ Hence giving me an answer of $ P(Z \leq 3,808,985,943)$ which definitely cannot be correct. Would appreciate it if somebody could point out my mistake.

Reattempt:

Using the hint i got i managed to deduce the following

$P(X_1 \cdot X_2 \cdot \ldots \cdot X_{30} \leq 1.85 \cdot 10^{-5})$

$P(log X_1 \cdot log X_2 \cdot \ldots \cdot log X_{30} \leq log 1.85 \cdot 10^{-5})$

P($\sum_{k=1}^{30} log X_i \leq log 1.85 \cdot 10^{-5})$

And since X random variable can be normally distributed X~ N( $\frac{1}{3}$ , $\frac{1}{9}$)

Substitute into the Z score

$ P(\frac{ log X -\mu }{\frac{\sigma}{\sqrt(n)}})\leq \frac { \frac{log1.85 \cdot 10^{-5}}{30} - (\frac{1}{3})}{(\frac{\frac{1}{3}}{\sqrt(n)})} $

$P(Z \leq 11.45)$

Am I on the right track? Is it correct to use $\mu$ = $\frac{1}{3}$ and $\sigma$ = $\frac{1}{3}$ in the z score?

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  • $\begingroup$ This question seems to be a self-study question. Please add the self-study tag if appropriate. $\endgroup$ – QuantIbex Apr 20 '14 at 13:40
  • $\begingroup$ What does the notation $P(X_1,X_2, \ldots, X_{30} \leq 1.85 \cdot 10^{-5})$ mean? Shouldn't it be $P(X_1 + X_2 + \cdots + X_{30} \leq 1.85 \cdot 10^{-5})$? $\endgroup$ – QuantIbex Apr 20 '14 at 14:00
  • $\begingroup$ My apologies Quantlbex, it suppose to be the product of the random variables. $P(X_1 \cdot X_2 \cdot X_3, \ldots,X_{30} \leq 1.85 \cdot 10^{-5})$ I apologies for the confusion $\endgroup$ – Ingrid Apr 20 '14 at 14:18
  • $\begingroup$ And how do we know that the product of i.i.d r.v.'s obeys the "usual" CLT? $\endgroup$ – Alecos Papadopoulos Apr 20 '14 at 14:40
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    $\begingroup$ Hint: $-\ln (X_1X_2\cdots X_n) = \sum_i -\ln(X_i) = \sum_i Y_i$ is a sum of exponential random variables $Y_i$. Can you relate $P(X_1X_2\cdots X_n \leq a)$ to $P\left(\sum_i Y_i \leq g(a)\right) = P\left(\sum_i Y_i \leq b\right)$ for some suitably chosen function $g(\cdot)$? $\endgroup$ – Dilip Sarwate Apr 20 '14 at 14:41
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The Op started correctly in the Re-attempt, but then lost it, so I provide the analytical steps. By the definition of the natural logarithm and its base, we have

$$X_i = \frac 1{e^{-\ln X_i}} \Rightarrow \prod_{i=1}^{30} X_i= \frac 1{\exp{\{\sum_{i=1}^{30}(-\ln X_i)}\}}$$

So

$$P(X_1 \cdot X_2 \cdot \ldots \cdot X_{30} \leq 1.85 \cdot 10^{-5}) = P\left(\frac 1{\exp{\{\sum_{i=1}^{30}(-\ln X_i)}\}} \leq 1.85 \cdot 10^{-5}\right)$$

$$=P\left(\frac {10^5}{1.85}\leq \exp{\{\sum_{i=1}^{30}(-\ln X_i)}\} \right)=P\left(\ln\left(\frac {10^5}{1.85}\right)\leq \sum_{i=1}^{30}(-\ln X_i) \right)$$

$$=1-P\left(\sum_{i=1}^{30}(-\ln X_i)\leq \ln\left(\frac {10^5}{1.85}\right) \right)$$

which is essentially where the OP arrived. But then the CLT is not to be used for one variable but for this sum of variables.

As already mentioned in the comments, this sum is the sum of i.i.d. exponential random variables, each having mean $1/3$ and variance $1/9$. So the sum, call it $S_n$, has mean and variance $E(S_n) = 30\frac 13 = 10$ and $\operatorname{Var}(S_n)= 30\frac 19 \Rightarrow \sigma_S = \frac {\sqrt{30}}{3}$

Then the normal approximation through the CLT tells us that $[S_n-E(S_n)]/\sigma_S \approx Z\sim N(0,1)$, and then numerical calculations.

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