# Confidence limit question

## Main Question or Discussion Point

Let f(x;p) = p*f(x) + (1-p)*g(x) where f(x) is the pdf of r.v. X1 dist N(1,1) and g(x) is the pdf of r.v. X2 dist N(0,1). Find one-sided lower confidence limit for p based on a sample size n=1.

This question has been driving me crazy. Everything that I've tried seems to be going nowhere, am I missing something obvious. What's throwing me off is that the new pdf is a sum of two others, and its not entirely obvious what the distribution f(x;p) looks like.

Thanks

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cZ~N(cmu,c^2sigma^2) where Z~N(mu,sigma^2)

and X+Y~N(mu_x+mu_y,sigma_x^2+sigma_y^2) (assuming x and y are independent).

thanks for the response,

just a clarification, I dont see how it's entirely obvious that c*f(x) is equivalent to cX. The latter is a pretty obvious result from MGF's but the former I'm not sure about.

Hmmm?

c*f_x(x) is not the density function for cX. Unless, c=1.

its alright, I have it.

To answer your question, thats exactly what I meant. cX is not equivalent to c*f(x)

But from you're earlier post, you just stated that a linear combination of normal variables is also normal. Thats only true when we're talking about the random variable itself, not its pdf. I was curious about the linear combination of the pdf of normal variables.

However, the answer doesn't depend on the distribution of f(x;p) anyways, so its all a moot point either way, just use the CDF transform technique and it becomes uniform(0,1).

Linear combinations of pdfs don't make sense unless a) they integrate to probability 1 and b) they are nonnegative.

cf(x) integreates to c whenever f(x) is a pdf.

thats pretty obvious, i don't see how it helps with the question, its figured out anyways