Random distribution

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Have no ideas how to start. Any thoughts?

Ivan Nikiforovich invests his savings in a mutual investment fund. The mutual fund yield is a random variable uniformly distributed over the interval [-10,20]% per annum. Until his retirement, Ivan Nikiforovich has 10 more years left. What is the probability of his savings during this time to grow at least twice?

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My idea is to calculate the average growth rate which lead to the double of the investment after 10 years. The equation is $(1+x)^{10}=2\Rightarrow x=2^{\frac1{10}}-1=0.0717734625362931$. With an mutual fund yield of at least $7.17735\%$ we can ensure that investment is at least twice after 10 years.

Next we have to calculate the probability that the mutual fund yield is greater than $7.17735\%$. Here we use the uniform distribution. $$P(X\geq 7.17735\%)=1-P(X\leq 7.17735\%)=1-\frac{0.0717735-(-0.1)}{0.2-(-0.1)}=0.4274217$$


Same approach with a slightly different perspective. We can just re-arrange the inequality. Basically the asked probability is $$P\left(\left(1+X \right)^{10}\geq 2\right),$$

where $X\sim U(-0.1,0.2)$

After some transformations we get

$$P\left(X \geq 2^{\frac1{10}}-1\right)$$

Thus Ivan Nikiforovich has a probability of $\color{blue}{42.74\%}$ that his savings grow at least twice in ten years.