Suppose I have a pdf $f(S)$. $f(S)$ describes the size of firms in the economy. Also define the Bernoulli variable $X_{f} \in \{0,1\}$ where $P(X_{f}=1)=g(S_{f})$ and $P(X_{f}=0)=1-g(S_{f})$. $S_{f}$ is the size of firm $f$ and $g(S_{f})$ depends positively on $S_{f}$.

Question: What does the firm size density look like for firms that have $X=1$? Is it the product of $f(S)$ and $g(S)$?

I think this is some kind of application of Poisson sampling where each element of the population that is sampled ($f(S)$) is subjected to an independent Bernoulli trial ($g(S)$) which determines whether the firm becomes part of the sample. I wonder what the density of this sample looks like.

For concreteness, suppose $f(S)\sim U[1,S_{max}]$, $S_{max}=100$ and $g(S_{f})=\frac{S_{f}}{S_{max}}$.

Any help is much appreciated.


Strictly speaking it appears you have been given $\mathsf P(X_f=1\mid S_f=S)=g(S)$.

The marginal probability of $X_f$ should not be a function of firm size (by definition).

$$\mathsf P(X_f=1) = \int_\Bbb R f(s)\cdot\mathsf P(X_f=1\mid S_f=s)\operatorname d s $$

You want to find $f(S\mid X_f=1) = \dfrac{f(S)\cdot\mathsf P(X_f=1\mid S_f=S)}{\int_{\Bbb R} f(s)\cdot\mathsf P(X_f=1\mid S_f=s)\operatorname d s} = \dfrac{f(S)\cdot g(S)}{\int_\Bbb R f(s)\cdot g(s)\operatorname d s }$

When $S_f\sim \mathcal U[1; 100],\; g(S_f) = \frac{S_f}{100}\mathbf 1_{S_f\in[1;100]}$ then

$$\begin{align}f(S\mid X_f=1) & = \dfrac{\cfrac 1{99} \cdot\cfrac{S}{100}}{\displaystyle\int_1^{100} \cfrac 1{99}\cfrac s{100}\operatorname d s} \;\mathbf 1_{S\in[1;100]}\\[2ex] ~ & = \dfrac{2S}{9999} \;\mathbf 1_{S\in[1;100]}\end{align}$$

  • $\begingroup$ Thank you very much for you reply. This has solved the issue. $\endgroup$
    – Rita V
    Sep 3 '15 at 6:52
  • $\begingroup$ @RitaV: If this answer resolved your question, you can mark it as accepted by clicking the green tick. $\endgroup$
    – user21820
    Mar 2 '16 at 9:33

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