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    <p>The problem with your function is that you have written it assuming that <code>x</code> is a single value, and then you have passed it a vector. Your code cannot cope with this.</p> <p>If you try to run your code with a single value, eg: <code>a=PdfMixUnfmBeta(15,0.4)</code> your function should run.</p> <p>Lets look at what is actually wrong with your code, when I tried running your code I was given the following error: </p> <pre><code>Output argument "y" (and maybe others) not assigned during call </code></pre> <p>This indicates that the lines which assign to <code>y</code> are never actually executed. This indicates that neither of the conditional statements (<code>x &lt; 0</code> and <code>x &gt;= 0</code>) ever evaluate as true. The <code>if</code> statements expects a scalar logical value, but in your example it is provided with a vector of logical. </p> <p>So, to fix this you either need to deal with your <code>x</code> values one at a time by wrapping it in a for loop, e.g.:</p> <pre><code>function y = PdfMixUnfmBeta(x_vec, alpha) for x = x_vec %do your function here end end </code></pre> <p>Alternatively, you can vectorize your code, which is by far the preferable solution:</p> <pre><code>y = zeros(1,length(x)); % where x is &lt; 0 use this formula y(x &lt; 0) = (1-alpha) .*(1/2); % where x is &gt;= 0 use this formula instead. y(x &gt;= 0) = (1-alpha) .* (1/2) + (alpha .* 6 .* x .* (1-x)); </code></pre> <p>In the above solution I use logical indexing and change some of the <code>*</code> to <code>.*</code>. You may find it useful to look up the difference between <code>*</code> and <code>.*</code> and to read up on vectorization.</p>
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    1. POWhat is wrong with this matlab function code?
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    1. COCool I got it. It works now
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