Probability - Exponential Distribution

Brains_Tom
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Here's the evil question:

Let X~Exponential(alpha). Derive and name the pdf of Y=(alpha)X
 
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Hi! You should show some of your thoughts or working in your post...

But anyway, here are some steps to guide you along.

Step 1: Find the cumulative distribution function (cdf) of X. Since X is continuous, you will need to integrate the pdf of X (f(x)=\alpha e^{-\alpha x}, for x\geq 0) , with the lower limit being 0 (since we define x\geq0 for an exponential distribution) and the upper limit an arbitrary constant x.

Step 2: Use the relation Y= \alpha X to derive the cdf of Y from the cdf of X. So F(y) = P(Y\leq y) = P(\alpha X\leq y) = P(X\leq \frac{y}{\alpha})

Step 3: We can calculate this final probability since we know the cdf of X.

Step 4: Finally, differentiate the cdf of Y to obtain its pdf.

You will get a nice answer in the end.

All the best!

Note: Letters in small casing (e.g. x, y) represent constants while block letters (e.g. X, Y) are used to define the random variables.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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